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https://mathoverflow.net/questions/10959 | 4 | I recently heard the following puzzle: There are three nails in the wall, and you want to hang a picture by wrapping a wire attached to the picture around the nails so that if any one nail is removed the picture still stays but if any two nails are removed then the picture falls down. An answer, schematically, is given by abca-1b-1c-1 (i.e., wrap it clockwise around each of the three nails in some order, then counterclockwise around each of the three nails in the same order).
This reminded me very much of the [Borromean rings](http://en.wikipedia.org/wiki/Borromean_rings%20%22Borromean%20rings%22), where three rings are linked but when any one ring is removed the other two become unlinked. So I was trying to figure out if there might be some way to transform one situation into the other. My first instinct was to put the rings in S3 and have one of them pass through ∞, but that isn't really right. What seems to be tripping me up is that with the picture there's an extra object (the wire) that doesn't show up with the Borromean rings, but I have a vague idea that perhaps we could change the former situation by saying that we make the loop in the complement of three unlinked rings, and then perhaps "pulling really hard on the wire" would somehow thread the rings together. Maybe my issue is that what's *really* going on with the picture is just that we're making a loop in the complement of three points in plane, and I'm confounding phenomena of different dimensions...
Does anyone have any ideas?
| https://mathoverflow.net/users/303 | relationship between borromean rings and hanging-a-picture-from-three-nails puzzle? | This question is related to this post and its answers:
[Collapsible group words](https://mathoverflow.net/questions/15316/collapsible-group-words)
You may think of the three nails as giving a 3-punctured
plane, which has fundamental group a rank 3 free group.
An element of this group may be thought of as pushing a point
around in the surface. If one takes the trace of this motion
in time, you get a braid with 4 strands, three strands of which
are straight. This represents an element of the pure braid group, an
example of the Birman exact sequence. Closing up the
braid, you get a four component link. Your condition implies
that removing any two of the last three strands gives the trivial link, so it is
a kind of second order Brunnian condition.
| 6 | https://mathoverflow.net/users/1345 | 18523 | 12,347 |
https://mathoverflow.net/questions/18514 | 5 | Given a topological space X, i'd like to find Der X - the derived category of sheaves of abelian groups on X - to be a closed monoidal category. Hom should be cohomological and the internal-hom should be triangulated.
1. Is this possible in full generality? (Unbounded complexes, no restrictions on X)
2. Consider a sheaf of rings R or equivalently a ring of sheaves. This gives us two things: An abelian category of left R modules that we can derive; let's call this one Der R; A monoid R in Der X whose category of modules we denote dMod R. Is Der R = dMod R'? Ff not: how do they relate?
3. Given monoids R,S,T in Der X do we get the usual adjunctions in two variables between their categories of bimodules?
4. Given rings R,S,T in Sh X do we get the usual adjunctions in two variables between their derived categories of bimodules?
Now for the question: What is the right setting to do this? As i understand it, there's no suitable model structure that gives 4 in full generality.
| https://mathoverflow.net/users/1261 | Closed monoidal structure on the derived category of sheaves | I think the natural context to answer this and related questions is that of symmetric monoidal stable $(\infty,1)$-categories. In that context Lurie's DAG I-III give all the necessary foundations to straightforwardly generalize all of the usual abelian story you refer to to the derived category. There is such an object for sheaves on X (unbounded complexes thereof), and we can talk about associative algebra objects therein, their stable module categories, bimodules etc. and everything works just as you'd expect. In particular the homotopy categories of all these will be what you expect. (If you're looking for a worked-out example of playing these games in a similar context, I can selfishly suggest [this](http://arxiv.org/abs/0805.0157).)
| 7 | https://mathoverflow.net/users/582 | 18524 | 12,348 |
https://mathoverflow.net/questions/18522 | 26 | So I did some algebraic topology at university, including homotopy theory and basic simplicial homology, as well as some differential geometry; and now I'm coming back to the subject for fun via Hatcher's textbook. A problem I had in the past and still have now is how to understand projective space RP^n - I just can't visualise it or think about it in any concrete way. Any ideas?
edit: Essentially RP^n is always the example I don't understand. So when for example Hatcher says that S^n is of a CW complex with two cells e^0 and e^n, I can picture what's going on because I know what spheres look like and I can imagine the attachment in some concrete-ish way. But when he says "we see that RP^n is obtained from RP^{n-1} by attaching an n-cell [...] it follows by induction on n that RP^n has a cell complex structure e^0 U e^1 U ... e^n" - my brain just gives up.
| https://mathoverflow.net/users/1256 | How should I visualise RP^n? | You can "visualize" the cell structure on $\mathbb{R}P^n$ rather explicitly as follows. The set of tuples $(x\_0, ... x\_n) \in \mathbb{R}^{n+1}$, not all equal to zero, under the equivalence relation where we identify two tuples that differ by multiplication by a nonzero real number, can be broken up into pieces depending on which of the $x\_i$ are equal to zero.
* If $x\_0 \neq 0$, the corresponding points can be written $(1, x\_1, ... x\_n)$, and they form a subspace isomorphic to $\mathbb{R}^n$.
* If $x\_0 = 0$ and $x\_1 \neq 0$, the corresponding points can be written $(0, 1, x\_2, ... x\_n)$, and they form a subspace isomorphic to $\mathbb{R}^{n-1}$.
And so forth. One way to say this is that the tuples where $x\_0 \neq 0$ form an affine slice or affine cover of $\mathbb{R}P^n$ and the tuples where $x\_0 = 0$ constitute the "points at infinity," which themselves form a copy of $\mathbb{R}P^{n-1}$.
| 19 | https://mathoverflow.net/users/290 | 18530 | 12,352 |
https://mathoverflow.net/questions/18464 | 5 | Definitions and the main question
---------------------------------
Recall that a category $\mathcal C$ is **monoidal** if it is equipped with the following data (two functors, three natural transformations, and some properties):
* a functor $\otimes: \mathcal C \times \mathcal C \to \mathcal C$,
* a functor $1: \{\text{pt}\} \to \mathcal C$,
* a natural transformation $\alpha: (X\otimes Y)\otimes Z \overset\sim\to X\otimes(Y\otimes Z)$ between functors $\mathcal C^{\times 3} \to \mathcal C$ (natural in $X,Y,Z$),
* natural transformations $\lambda: 1 \otimes X \overset\sim\to X$, $\rho: X\otimes 1 \overset\sim\to X$ between functors $\mathcal C \to \mathcal C$ (natural in $X$; this uses the canonical isomorphisms of categories $\{\text{pt}\} \times \mathcal C \cong \mathcal C \cong \mathcal C \times \{\text{pt}\}$),
* such that $\alpha$ satisfies a pentagon,
* and $\alpha,\lambda,\rho$ satisfy some other equations.
I tend to be less interested in the unit laws $\lambda,\rho$, which is my excuse for knowing less about their technicalities. In my experience, it's the associativity law $\alpha$ that can have interesting behavior.
Let $\mathcal C,\mathcal D$ be monoidal categories. Recall that a functor $F: \mathcal C \to \mathcal D$ is **(strong) monoidal** if it comes with the following data (two natural transformations, and three properties):
* a natural isomorphism $\phi: \otimes\_{\mathcal D} \circ (F\times F) \overset\sim\to F\circ \otimes\_{\mathcal C}$ of functors $\mathcal C \times \mathcal C \to \mathcal D$,
* a natural isomorphism $\varphi: 1\\_{\mathcal D} \overset\sim\to F\circ 1\\_{\mathcal C}$ of functors $\{\text{pt}\} \to \mathcal D$,
* satisfying some properties, the main one being that the two natural transformations $(FX \otimes\_{\mathcal D} FY) \otimes\_{\mathcal D} FZ \overset\sim\to F(X\otimes\_{\mathcal C} (Y \otimes\_{\mathcal C}Z))$ that are built from $\phi, \alpha\_{\mathcal C}, \alpha\_{\mathcal D}$ agree. This property expresses that the associators in $\mathcal C$, $\mathcal D$ are "the same" under the functor $F$.
>
> My question is whether there is a (useful) weakening of the axioms for a monoidal functor that expresses the possibility that the associators might disagree.
>
>
>
An example: quasiHopf algebras
------------------------------
Here is my motivating example. Let $A$ be a (unital, associative) algebra (over a field $\mathbb K$), and let $A\text{-rep}$ be its category of representations. I.e. objects are pairs $V \in \text{Vect}\\_{\mathbb K}$ and an algebra homomorphism $\pi\_V: A \to \text{End}\\_{\mathbb K}(V)$, and morphisms are $A$-linear maps. Then $A\text{-rep}$ has a faithful functor $A\text{-rep} \to \text{Vect}\\_{\mathbb K}$ that "forgets" the map $\pi$.
Suppose now that $A$ comes equipped with an algebra homomorphism $\Delta: A \to A \otimes\_{\mathbb K} A$. Then $A\text{-rep}$ has a functor $\otimes: A\text{-rep} \times A\text{-rep} \to A\text{-rep}$, given by $\pi\_{(V\otimes W)} = (\pi\_V \otimes \pi\_W) \circ \Delta: A \to \text{End}(V\otimes\_{\mathbb K}W)$. Just this much data is not enough for $A\text{-rep}$ to be monoidal. (Well, we also need a map $\epsilon: A \to \mathbb K$, but I'm going to drop mention of the unit laws.) Indeed: there might not be an associator.
A situation in which there *is* an associator on $(A\text{-rep},\otimes)$ is as follows. Suppose that there is an invertible element $p \in A^{\otimes 3}$, such that for each $a\in A$, we have
$$ p\cdot (\Delta \otimes \text{id})(\Delta(a)) = (\text{id} \otimes \Delta)(\Delta(a))\cdot p $$
and $\cdot$ is the multiplication in $A^{\otimes 3}$. Then for objects $(X,\pi\_X), (Y,\pi\_Y), (Z,\pi\_Z) \in A\text{-rep}$, define:
$$ \alpha\_{X,Y,Z} = (\pi\_X \otimes \pi\_Y \otimes \pi\_Z)(p) :
((X\otimes\\_{A\text{-}{\rm rep}} Y) \otimes\\_{A\text{-}{\rm rep}} Z) \to (X \otimes\\_{A\text{-}{\rm rep}} (Y \otimes\\_{A\text{-}{\rm rep}} Z)) $$
You can check that it is in fact a isomorphism in $A\text{-rep}$. Moreover, supposing that $p$ satisfies:
$$ (\text{id} \otimes \text{id} \otimes \Delta)(p) \cdot (\Delta \otimes \text{id} \otimes \text{id})(p) = (1 \otimes p) \cdot (\text{id} \otimes \Delta \otimes \text{id})(p) \cdot (p \otimes 1) $$
where now $\cdot$ is the multiplication in $A^{\otimes 4}$, then $\alpha$ is an honest associator on $A\text{-rep}$.
Then (provided also that $A$ have some sort of "antipode"), the data $(A,\Delta,p)$ is a **quasiHopf algebra**.
Anyway, it's clear from the construction that the forgetful map $\text{Forget}: A\text{-rep} \to \text{Vect}\\_{\mathbb K}$ is a faithful exact functor which is weakly monoidal in the sense that $\text{Forget}(X \otimes\\_{A\text{-}{\rm rep}} Y) = \text{Forget}(X) \otimes\\_{\mathbb K} \text{Forget}(Y)$ — indeed, this is equality of objects, so perhaps it is "strictly" monoidal — but it is not "monoidal" since it messes with the associators.
Actual motivation
-----------------
My actual motivation for asking the question above is the understand the Tannaka duality for quasiHopf algebras. In general, we have the following theorem:
**Theorem:** Let $\mathcal C$ be an abelian category and $F: \mathcal C \to \text{FinVect}\\_{\mathbb K}$ a faithful exact functor, where $\text{FinVect}\\_{\mathbb K}$ is the category of finite-dimensional vector spaces of $\mathbb K$. Then there is a canonical coalgebra $\text{End}^{\vee}(F)$, and $\mathcal C$ is equivalent as an abelian category to the category of finite-dimensional corepresentations of $\text{End}^{\vee}(F)$.
For details, see A Joyal, R Street, An introduction to Tannaka duality and quantum groups, Category Theory, Lecture Notes in Math, 1991 vol. 1488 pp. 412–492.
The Tannaka philosophy goes on to say that if in addition to the conditions in the theorem, $\mathcal C$ is a monoidal category and $F$ is a monoidal functor, then $\text{End}^{\vee}(F)$ is a bialgebra, and $\mathcal C$ is monoidally equivalent to $\text{End}^{\vee}(F)\text{-corep}$. If $\mathcal C$ has duals, $\text{End}^{\vee}(F)$ is a Hopf algebra. If $\mathcal C$ has a braiding, then $\text{End}^{\vee}(F)$ is coquasitriangular. Etc.
My real question, then, is:
>
> What is the statement for Tannaka duality for (co)quasiHopf algebras?
>
>
>
It seems that the standard paper to answer the real question is:
S. Majid, Tannaka-Krein theorems for quasi-Hopf algebras and other results. Contemp. Math. 134 (1992), pp. 219–232.
But I have not been able to find a copy of this paper yet.
| https://mathoverflow.net/users/78 | What are the correct axioms for a "weakly associative monoidal functor"? | Theo,
I believe the notion that you're looking for is called quasi-fiber functor, and is discussed in:
<http://www-math.mit.edu/~etingof/tenscat1.pdf>
It is also discussed how to do Tannakian formalism in the presence of a quass-fiber functor, and you get back a quasi-Hopf algebra, as you desire. I'll omit repeating an explanation here, since it's given nicely in the appropriate section linked.
-david
| 3 | https://mathoverflow.net/users/1040 | 18536 | 12,355 |
https://mathoverflow.net/questions/12068 | 17 | Let $X$ be a smooth surface of genus $g$ and $S^nX$ its n-symmetrical product (that is, the quotient of $X \times ... \times X$ by the symmetric group $S\_n$). There is a well known, cool formula computing the Euler characteristic of all these n-symmetrical products:
$$\sum\_{d \geq 0} \ \chi \left(X^{[d]} \right)q^d \ \ = \ \ (1-q)^{- \chi(X)}$$
It is known that $S^nX \cong X^{[n]}$, the Hilbert scheme of 0-subschemes of length n over $X$. Hence, the previous formula also computes the Euler characteristic of these spaces.
What about for singular surfaces? More precisely, if $X$ is a singular complex algebraic curve, do you know how to compute the Euler characteristic of its n-symmetrical powers $S^nX$? More importantly: what is the Euler characteristic of $X^{[n]}$, the Hilbert scheme of 0-schemes of length n over $X$?
I guess it is too much to hope for a formula as neat as the one given for the smooth case. Examples, formulas for a few cases or general behaviour (e.g. if for large n, $\chi\left(X^{[n]}\right) = 0)$ are all very welcome!
| https://mathoverflow.net/users/3314 | What is the Euler characteristic of a Hilbert scheme of points of a singular algebraic curve? | For singular plane curves, there is a conjectural formula (due to Alexei Oblomkov and myself) in terms of the HOMFLY polynomial of the links of the singularities. For curves whose singularities are torus knots, i.e. like x^a = y^b for a,b relatively prime, and for a few more singularities, the conjecture has been established. See [this preprint](https://arxiv.org/abs/1003.1568).
**Edit:** More recently I have given [a different characterization](https://arxiv.org/abs/1009.0914) of these numbers in terms of multiplicities of certain strata in the versal deformation of the singular curve.
| 16 | https://mathoverflow.net/users/4707 | 18540 | 12,357 |
https://mathoverflow.net/questions/18544 | 13 | Is there a good way to define a sheaf over a simplicial set - i.e. as a functor from the diagram of the simplicial set to wherever the sheaf takes its values - in a way that while defined on simplex by simplex corresponds in some natural manner to what a sheaf over the geometric realization of the simplicial set would look like?
*Edited to add:* I'm actually interested in the question in a concrete fashion rather than an abstract one - I'm trying to figure out whether there might be some interesting interpretation of sheaves over the nerve of a category generated by a network; similar to current work by Robert Ghrist that takes a network and views it as a graph, and thus as a topological space (1-dim simplicial complex), and manages to find useful interpretations of sheaves on this particular space in terms of network analysis.
Hence, what I'm really looking for is an interesting definition for, say, the nerve of the category generated by a finite directed graph, or so...
*Edited to add:* In off-channels, fpqc has clarified his argument in the answer I've accepted. Specifically, $N(C)$ for a category has inherent direction data that is lost in $|N(C)|$. This messes up attempts to formulate an idea of *sheaves* over $N(C)$ in a way that stays faithful to the definition of sheaves over $|N(C)|$.
| https://mathoverflow.net/users/102 | Sheaves over simplicial sets | Clearly looking at sheaves on the geometric realisation gives something too far
removed from the simplicial picture. This is essentially because there are too
many sheaves on a simplex have (most of which are unrelated to simplicial
ideas). What one could do is to consider such sheaves which are constructible
with respect to the skeleton filtration, i.e., are constant on each open
simplex. This can be described inductively using Artin gluing. I think it
amounts to the following for a simplicial set $F$.
For each simplex $c\in F\_n$ we have a set $T\_c$, the constant value of the sheaf
$T$ on the interior of the simplex corresponding to $c$.
For each surjective map $f\colon [n] \to [m]$ in $\Delta$ the corresponding (degeneracy) map
on geometric simplices maps the interior of $\Delta\_n$ into (onto in fact) the interior of $\Delta\_m$ and
hence we have a bijection $T\_{f(c)} \to T\_c$. These bijections are transitive with
respect to compositions of $f$'s.
For each injective map $f\colon [m] \to [n]$ in $\Delta$ the corresponding map on geometric
simplices maps $\Delta\_m$ onto a closed subset of $\Delta\_n$. If $j\colon \Delta^o\_n \hookrightarrow \Delta\_n$ is the
inclusion of the interior we get an adjunction map $T \to j\_\ast j^\ast T$ and
$j\_\ast j^\ast T=T\_c$ where $T\_c$ also denotes the constanct sheaf with value $T\_c$. If
$f'\colon \Delta^o\_m \hookrightarrow \Delta\_n$ is the inclusion of the interior composed with $f$ we can
restrict the adjunction map to get a map $T\_{f(c)}=f'^\ast T \to f'T\_c$ and taking
global sections we get an actual map $T\_{f(c)} \to T\_c$. These maps are transitive
with respect compositions of $f$'s.
We have a compatibility between maps coming from surjections and
injections. Unless something very funny is going on this compatibility should be
that we wind up with a function on the comma category $\Delta/F$ which takes
surjections $[n] \to [m]$ to isomorphisms.
There is the stronger condition on the sheaf $F$, namely that it is constant on
each star of each simplex. This means on the one hand that it is locally
constant on the geometric realisation, on the other hand that $T\_{f(c)} \to T\_c$
is always an isomorpism.
[Added] Some comments intended to give some kind of relation with the answer
provided by fpqc. My suggested answer is not homotopy invariant in the sense
that a weak (or even homotopy) equivalence of simplicial sets does not induce an
equivalence on the category of sheaves. This is so however if one, as per above,
adds the condition that all the maps $T\_{f(c)} \to T\_c$ are isomorphisms. However,
that condition is not so good as many maps that are not weak equivalences
induces category equivalences (it is enough that the map induce isomorphisms on
$\pi\_0$ and $\pi\_1$). This is a well-known phenomenon and has to do with the fact
the $T\_c$ are just sets. One could go further and assume that the $T\_c$ are
topological spaces and the maps $T\_{f(c)} \to T\_c$ continuous. Of course adding
the condition that these maps be homeomorphisms shouldn't be right thing to do,
instead one should demand that they be homotopy (or weak) equivalences. Again,
this shouldn't be quite it because of the transitivity conditions. We should not
have that the composite $T\_{g(f(c))} \to T\_{f(c)} \to T\_c$ should be equal to
$T\_{g(f(c))} \to T\_c$ but rather homotopic to it. Once we have opened that can of
worms we should impose higher homotopies between repeated composites. This can
no doubt be (has been) done but there seems to be an easier way out. In the
first step away from set-valued $T\_c$ we have the possibility of they being
instead categories. In that case the higher homotopy conditions is that we
should have a pseudofunctor $\Delta/F \to \mathcal{C}\mathrm{at}$. Even they are
somewhat unpleasant and it is much better to pass to the associated fibred
category $\mathcal{T} \to \Delta/F$. In the general case, and admitting that $\Delta/F$ is
essentially the same things as $F$ itself, we should therefore look at
(Serre) fibrations $X \to |F|$ or if we want to stay completely simplicial, Kan
fibrations $X \to F$. This gives another notion of (very flabby) sheaf which now
should be homotopy invariant (though that should probably be in the sense of
homotopy equivalence of $\Delta$-enriched categories).
| 10 | https://mathoverflow.net/users/4008 | 18549 | 12,361 |
https://mathoverflow.net/questions/18547 | 18 | I'm interested in bounds for the number of unique determinants of NxN (0,1)-matrices. Obviously some of these matrices will be singular and therefore will trivially have zero determinant. While it might also be interesting to ask what number of NxN (0,1)-matrices are singular or non-singular, I'd like to ignore singular matrices altogether in this question.
To get a better grasp on the problem I wrote a computer program to search for the values given an input N. The output is below:
1x1: 2 possible determinants
2x2: 3 ...
3x3: 5 ...
4x4: 9 ...
5x5: 19 ...
Because the program is simply designed to just a brute force over every possible matrix the computation time grows with respect to $O(2^{N^2})$. Computing 6x6 looks like it is going to take me close to a month and 7x7 is beyond hope without access to a cluster. I don't feel like this limited amount of output is enough to make a solid conjecture.
I have a practical application in mind, but I'd also like to get the bounds to satiate my curiosity.
| https://mathoverflow.net/users/3737 | Number of unique determinants for an NxN (0,1)-matrix | I have given some detail in a comment to another answer. I have a proof that the number of determinants is greater than 4 times the nth Fibonacci number for (n+1)x(n+1) (0,1) matrices, and I conjecture that for large n the number of distinct determinants approaches a constant times n^(n/2). Math Overflow has some hints of the proof in answers I made on other questions.
I am interested in your idea for an application, and am willing to share more information on this subject.
Gerhard "Ask Me About System Design" Paseman, 2010.03.19
| 4 | https://mathoverflow.net/users/3206 | 18554 | 12,364 |
https://mathoverflow.net/questions/18553 | 5 | Let $S\_4 = \left(\begin{array}{cc}0&-1 \\\ 1&0 \end{array}\right) \textrm{ and } S\_6 = \left(\begin{array}{cc} 1&-1 \\\ 1&0\end{array}\right)$. Serre proves in his book on trees that $SL\_2(\mathbb{Z}) \cong \mathbb{Z}/4 \*\_{\mathbb{Z}/2} \mathbb{Z}/6$, and $S\_4$ and $S\_6$ are the elements corresponding to the generators of $\mathbb Z/4$ and $\mathbb Z/6$ (I'm not sure if this is related to my question). Then let $a = S\_4 S\_6$ and $b = S\_4 S\_6^2$. I believe every element of $SL\_2(\mathbb Z)$ can be written as $S\_6^d w S\_6^e$, where $w$ is a word in $a$ and $b$ but not $a^{-1}$ or $b^{-1}$.
I wrote a program (for other purposes) that seems to show that there aren't any relations between $a$ and $b$ that have length 15 or less and don't involve $a^{-1}$ or $b^{-1}$. I'm not certain that the program is right, but if it is, one might make a naive guess that these two elements generate a free group. This makes me suspicious.
1) Does $SL\_2(\mathbb Z)$ contain a free group (of rank > 1)? If it does, is there an easy way to determine whether the subgroup generated by $a$ and $b$ is free?
2) A slightly less naive guess is that $a$ and $b$ generate a free monoid in $SL\_2(\mathbb Z)$. Is there a reason why $SL\_2(\mathbb Z)$ can't contain a free monoid, or an example showing that it does?
EDIT: Thanks for the quick replies. As Robin and Jack pointed out, $a$ and $b$ generate SL(2,Z), so clearly don't generate a free group. Also, there are free subgroups that are easy to write down. I'm still curious about #2, though.
| https://mathoverflow.net/users/2669 | Relations between two particular elements of SL_2(Z)? | Certainly $\mathrm{SL}(2,\mathbb{Z})$ contains a free group.
For instance $\Gamma(2)$, the subgroup of all matrices congruent
to the identity modulo $2$, is free of rank $2$. The matrices
$\left(\begin{array}{cc}1&2\\\ 0&1\end{array}\right)$
and
$\left(\begin{array}{cc}1&0\\\ 2&1\end{array}\right)$
freely generate $\Gamma(2)$.
This can be proved by considering the action on the upper half-plane
or by careful examination of reduced words.
There's a nice proof in chapter 18 of David Ullrich's
book *Complex Made Simple*.
Your $a$ and $b$ don't generate a free group alas, since they
generate all of $\mathrm{SL}(2,\mathbb{Z})$.
Re the edited question. Let's write
$$T=\left(\begin{array}{cc}1&1\\\ 0&1\end{array}\right)\qquad
\textrm{and}\qquad
U=\left(\begin{array}{cc}1&0\\\ 1&1\end{array}\right).$$
As both Jack and I pointed out, $T^2$ and $U^2$ generate
a free subgroup of rank $2$. Now it's an easy exercise to prove that
$T$ and $U$ freely generate a free monoid of rank $2$ (because
their entries are non-negative). On the other hand, they generate
thw whole group $\mathrm{SL}(2,\mathbb{Z})$ which is certainly
not free. Your matrices $a$ and $b$ are, if my calculations
are right, $-U^{-1}$ and $-T^{-1}$. The matrix $S\_4$ conjugates $T$
and $U$ into $U^{-1}$ and $T^{-1}$ so $U^{-1}$ and $T^{-1}$ freely
generate a free monoid of rank $2$. The same must be true of
$U^{-1}$ and $T^{-1}$, that is, of $a$ and $b$.
| 11 | https://mathoverflow.net/users/4213 | 18556 | 12,366 |
https://mathoverflow.net/questions/18539 | 19 | As I understand it, [Lions and DiPerna demonstrated existence and uniqueness](http://www.jstor.org/pss/1971423) for the [Boltzmann equation](http://en.wikipedia.org/wiki/Boltzmann_equation). Moreover, [this paper](http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6VJ2-44D2CY3-J&_user=10&_coverDate=11%252F30%252F2001&_rdoc=1&_fmt=high&_orig=search&_sort=d&_docanchor=&view=c&_searchStrId=1254807045&_rerunOrigin=google&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=e923283d5d59dec6efcfaab5d995390c) claims that
>
> Appropriately scaled families of
> DiPerna–Lions renormalized solutions
> of the Boltzmann equation are shown to
> have fluctuations whose limit points
> (in the weak $L^1$ topology) are governed
> by a Leray solution of the limiting
> Navier–Stokes equations.
>
>
>
Probably there is a lot of other work along these lines. But I am not well-versed enough in these areas to go through the literature easily, and so I hope someone can give a very high-level answer to my question:
>
> Why does renormalizing the Boltzmann
> equation not (yet?) give existence and
> uniqueness for Navier-Stokes?
>
>
>
| https://mathoverflow.net/users/1847 | Why don't existence and uniqueness for the Boltzmann equation imply the same for Navier-Stokes? | Okay, after figuring out which paper you were trying to link to in the third link, I decided that it is better to just give an answer rather then a bunch of comments. So... there are several issues at large in your question. I hope I can address at least some of them.
The "big picture" problem you are implicitly getting at is the Hilbert problem of hydrodynamical limit of the Boltzmann equations: that intuitively the ensemble behaviour at the large, as model by a fluid as a vector field on a continuum, should be derivable from the individual behaviour of particles, as described by kinetic theory. Very loosely tied to this is the problem of global existence and regularity of Navier-Stokes.
If your goal is to solve the Navier-Stokes problem using the hydrodynamic limit, then you need to show that (a) there are globally unique classical solutions to the the Boltzmann equations and (b) that they converge in a suitably regular norm, in some rescaling limit, to a solution of Navier-Stokes. Neither step is anywhere close to being done.
As far as I know, there are no large data, globally unique, classical solutions to the Boltzmann equation. Period. If we drop some of the conditions, then yes: for small data (perturbation of Maxwellian), the recent work of Gressman and Strain (0912.0888) and Ukai et al (0912.1426) solve the problem for long-range interactions (so not all collision kernels are available). If you drop the criterion of global, there are quite a bit of old literature on local solutions, and if you drop the criterion of unique and classical, you have the DiPerna-Lions solutions (which also imposes an angular-cutoff condition that is not completely physical).
The work of Golse and Saint-Raymond that you linked to establishes the following: that the weak solution of DiPerna-Lions weakly converges to the well-known weak solutions of Leray for the Navier-Stokes problem. While this, in some sense, solve the problem of Hilbert, it is rather hopeless for a scheme trying to show global properties of Navier-Stokes: the class of Leray solutions are non-unique.
As I see it, to go down this route, you'd need to (i) prove an analogue of DiPerna-Lions, or to get around it completely differently, and arrive at global classical and unique solutions for Boltzmann. This is a difficult problem, but I was told that a lot of very good people are working on it. Then you'd need (ii) also to prove an analogue of Golse-Saint-Raymond in a stronger topology, or you can use Golse-Saint-Raymond to first obtain a weak-limit that is a Leray solution, and then show somehow that regularity is preserved under this limiting process. This second step is also rather formidable.
I hope this somewhat answers your question.
| 22 | https://mathoverflow.net/users/3948 | 18567 | 12,371 |
https://mathoverflow.net/questions/18551 | 4 | It follows from a [recent answer](https://mathoverflow.net/questions/16132/formally-etale-at-all-primes-does-not-imply-formally-etale/17775#17775) that even when a ring is formally étale rather than étale, we can check this condition on localizations, and hence stalks. It's not hard to show that we can define a "formally étale" topology on $Aff$. Presumably there's a good reason for requiring finite presentation), but I can't think of a reason why off of the top of my head.
Specifically, we let the covering families be jointly faithfully flat with each morphism formally étale. This is by construction subcanonical.
Questions: Why are finiteness conditions necessary for schemes in general and étale maps in particular.
What kinds of problems will we run into if we do not put finitness conditions on the "formally étale" topology?
If this topology fails in some way, can requiring that covers are finite families of morphisms (quasicompact), or that the morphisms in the cover are themselves flat, or even both? This would give us a topology similar to the fpqc topology, except in that all covering families would be made up of formally étale morphisms. The only difference between this topology and the étale topology is the finite presentation of the morphisms in the covering families. Does this still not work?
| https://mathoverflow.net/users/1353 | Why do we need finiteness conditions for formally étale morphisms? | I guess, to get a good theory, you would want to be able to have effective descent for certain classes maps. In Brian Conrads answer to this question:
"[Quasi-separatedness for Algebraic Spaces](https://mathoverflow.net/questions/16381/quasi-separatedness-for-algebraic-spaces)"
he gives references to how you can work with algebraic spaces without separated assumptions. The cruical step in proving that étale equivalence relations give schematic quotient maps, involves descending étale separated maps. This is possible (in the affine case) for faithfully flat, universally open maps (i.e étale or fppf). Even if you set up your topology such that it lies between the fpqc and Zariski topologies, it is not immediate that this would work.
By the way, the only counterexamples I have seen for formaly étale maps not being étale, involves non flat maps. This leads to two questions:
* Are there flat, formally étale maps (between affines) which aren't fppf?
* Are there flat, formally étale, universally open maps (between affines) which aren't fppf?
| 2 | https://mathoverflow.net/users/1084 | 18587 | 12,385 |
https://mathoverflow.net/questions/18429 | 9 | Following the first chapter of Hatcher's great book "Spectral Sequences in Algebraic Topology", I got into problems with spectral sequences of cohomological type. Fix a ring $R$ once and for all. Please let me first recapitulate the **homological situation**.
An *exact couple* consists of bigraded $R$-modules $A$ and $E$ and bigraded $R$-module homomorphisms $i$, $j$, and $k$, such that
$$
\begin{array}{rcl}A&\xrightarrow{i}&A\newline {\scriptsize k}\nwarrow&&\swarrow{\scriptsize j}\newline&E&\end{array}
$$
is exact at every corner. Its derived pair with $A'=i(A)$ and $E'=H(E)$ with respect to $d=j\circ k$ is again exact. Before you wonder about the indices in what comes next, please continue reading up to the canonical example. This will explain the degrees, if I have not made a mistake. Let $a\in \mathbb{N}$ (in the example below we have $a=1$). Set $r=0$ for the moment. An exact couple with bidegrees
$$
\begin{array}{rcl}A^{(r)}&\xrightarrow{(1,-1)}&A^{(r)}\newline {\scriptsize (-1,0)}\nwarrow&&\swarrow{\scriptsize (-(a-1+r),(a-1+r))}\newline&E^{(r)}&\end{array}
$$
induces a spectral sequence $E\_{p,q}^{r+a}=E\_{p,q}^{(r)}$ of homological type with differentials $d^{r+a}=j^{(r)}\circ k^{(r)}$. Here ${\scriptsize something}^{(r)}$ denotes the corresponding term in the $r$-th derived couple. The $r$-th derived couple has bidegrees as in the diagram above.
Here is the canonical example. Let $0=C\_{-1}\subseteq C\_0\subseteq\ldots C\_p\subseteq\ldots =C$ be a filtration of a chain complex $C$. Take for example the singular complex of a topological space $X$ filtered by a filtration of $X$. We have many long exact sequences in homology, here are three of them:
$$
\begin{array}{lcccccr}
\to H\_{p+q}(C\_p,C\_{p-1})&\xrightarrow{k}&H\_{p+q-1}(C\_{p-1})&\xrightarrow{i}&H\_{p+q-1}(C\_p)&\xrightarrow{j}&H\_{p+q-1}(C\_p,C\_{p-1})\to\newline
\to H\_{p+q}(C\_{p-1},C\_{p-2})&\xrightarrow{k}&H\_{p+q-1}(C\_{p-2})&\xrightarrow{i}&H\_{p+q-1}(C\_{p-1})&\xrightarrow{j}&H\_{p+q-1}(C\_{p-1},C\_{p-2})\to\newline
\to H\_{p+q}(C\_{p-2},C\_{p-3})&\xrightarrow{k}&H\_{p+q-1}(C\_{p-3})&\xrightarrow{i}&H\_{p+q-1}(C\_{p-2})&\xrightarrow{j}&H\_{p+q-1}(C\_{p-2},C\_{p-3})\to
\end{array}
$$
There is an exact couple as above with $a=1$ and $E\_{p,q}=H\_{p+q}(C\_p,C\_{p-1})$ and $A\_{p,q}=H\_{p+q}(C\_p)$. Please note, that all the bigrades are correct. To follow $d^1:E\_{p,q}^1\to E\_{p-1,q}^1$, you start at the upper left corner, apply $k$, go one row down (the entries are equal here if you move one right) and apply $j$. One can also follow $d^2$ but this is not quite correct since you have to deal with representatives in $E^1$. Here you start at the upper left corner and get to the lower right.
Here is Hatcher's illuminating argument for convergence. I have never seen this so clearly presented.
>
> The spectral sequence $E^1(C\_\bullet)$ of homological type
> converges to $H\_{p+q}(C)$ if it is bounded (=only
> finitely many non-zero entries on
> every fixed diagonal $p+q$). One has
> $$ E^\infty\_{p,q}=i(H\_{p+q}(C\_p))/i(H\_{p+q}(C\_{p-1})) .$$
> ($i$ denotes the image in the colimit $H\_{p+q}(C)$.)
>
>
>
*Proof.* Let $r$ be large and consider (I don't know if this renders correctly) the $r$-th derived couple
$$
\begin{array}{c}
\to E^{r}\_{p+r,q-r+1}\xrightarrow{k^{r}} A^{r}\_{p+r-1,q-r+1}\xrightarrow{i^{r}}A^{r}\_{p+r,q-r}\newline
\xrightarrow{j^{r}}E^{r}\_{p,q}\xrightarrow{k^{r}}\newline
A^{r}\_{p-1,q}\xrightarrow{i^{r}}A^{r}\_{p,q-1}\xrightarrow{j^{r}}E^{r}\_{p-r,q-1+r}\to.
\end{array}
$$
The first and the last term are $0$ because of the bounding assumption. We have $A\_{p,q}^{r}=i(A^1\_{p-r,q+r})$ and since $C\_n$ is zero for $n<0$, the last three terms are $0$. Exactness implies the result.
Now the **cohomological situation**.
Let $0=C\_{-1}\subseteq C\_0\subseteq\ldots C\_p\subseteq\ldots =C$ be a filtration of a chain complex $C$. Take again for example the singular complex of a topological space $X$ filtered by a filtration of $X$. Understand the cohomology $H^n(C)$ of $C$ as $H\_n(\hom(C,\mathbb{Z}))$, the homology of the dualized complex as one does in topology.
Again we have many long exact sequences:
$$
\begin{array}{lcccccr}
\to H^{p+q}(C\_p,C\_{p-1})&\xrightarrow{k}&H^{p+q}(C\_p)&\xrightarrow{i}&H^{p+q}(C\_{p-1})&\xrightarrow{j}&H^{p+q+1}(C\_p,C\_{p-1})\to\newline
\to H^{p+q}(C\_{p+1},C\_p)&\xrightarrow{k}&H^{p+q}(C\_{p+1})&\xrightarrow{i}&H^{p+q}(C\_{p})&\xrightarrow{j}&H^{p+q+1}(C\_{p+1},C\_{p})\to\newline
\to H^{p+q}(C\_{p+2},C\_{p+1})&\xrightarrow{k}&H^{p+q}(C\_{p+2})&\xrightarrow{i}&H^{p+q}(C\_{p+1})&\xrightarrow{j}&H^{p+q+1}(C\_{p+2},C\_{p+1})\to
\end{array}
$$
One can again follow $d^1:E\_{p,q}^1\to E\_{p+1,q}^1$, etc. What is an exact couple of cohomological type? The only thing which fits into the picture is this: Set $A^{p,q}=H^{p+q}(C\_p)$ and $E^{p,q}=H^{p+q}(C\_p,C\_{p-1})$ and $a=1$.
An exact couple with bidegrees
$$
\begin{array}{rcl}A\_{(r)}&\xrightarrow{(-1,1)}&A\_{(r)}\newline {\scriptsize (0,0)}\nwarrow&&\swarrow{\scriptsize (a+r,-(a-1+r))}\newline&E\_{(r)}&\end{array}
$$
induces a spectral sequence $E\_{r+a}=E\_{(r)}$ of cohomological type with differentials $d\_{r+a}=j\_{(r)}\circ k\_{(r)}$. Here ${\scriptsize something}\_{(r)}$ denotes the corresponding term in the $r$-th derived couple. The $r$-th derived couple has bidegrees as in the diagram.
Now I would like to establish the following result. This is done nowhere since everything is said to be "dual" to the homological case. But if you look at the indices...hmm:
>
> The spectral sequence $E\_1(C\_\bullet)$ of cohomological type
> converges to $H^{p+q}(C)$ if it is bounded. One has
> $$ E\_\infty^{p,q}=ker(H^{p+q}(C)\to H^{p+q}(C\_{p-1}))/ker(H^{p+q}(C)\to H^{p+q}(C\_{p})) .$$
>
>
>
*Proof?* Let $r$ be large and consider (I don't know if this renders correctly) the $r$-th derived couple
$$
\begin{array}{c}
\to E\_{r}^{p-r,q+r-1}\xrightarrow{k\_{r}} A\_{r}^{p-r,q+r-1}\xrightarrow{i\_{r}}A\_{r}^{p-r-1,q+r}\newline
\xrightarrow{j\_{r}}E\_{r}^{p,q}\xrightarrow{k\_{r}}\newline
A\_{r}^{p,q}\xrightarrow{i\_{r}}A\_{r}^{p-1,q+1}\xrightarrow{j\_{r}}E\_{r}^{p+r,q-r+1}\to.
\end{array}
$$
The last term is $0$ because of the bounding assumption. We have $A\_{r}^{p,q}=i(A^{p+r,q-r}\_1)$ and since $C\_n$ is zero for $n<0$, the second and the third term are $0$. Exactness implies that
$$
E\_r^{p,q}=ker(i(H^{p+q}(C\_{p+r}))\to i(H^{p+q}(C\_{p+r-1}))).
$$
I cannot see how the result follows from this. Perhaps it is "only" a limit trick but I can not see it. So the question is:
>
> *How does Lemma 1.2. of Hatcher's text works in the cohomological case?*
>
>
>
| https://mathoverflow.net/users/4676 | Convergence of spectral sequences of cohomological type | The lemma you refer to has two halves. The first half covers the case of homology and the second half covers the case of cohomology. Proofs are given for both halves (though the last sentence of the proof in the cohomology case requires a moment's thought to convince oneself of). The way to derive the cohomology spectral sequence from the second half is explained on the page following the lemma. Perhaps there is something that seems unclear in what is written there? Please feel free to ask me about this, though this might be better done in email.
| 11 | https://mathoverflow.net/users/23571 | 18590 | 12,387 |
https://mathoverflow.net/questions/18438 | 2 | Given a set of distances S, choose N unique points P on a number line such that the distances between the N points occur in S as much as possible. That is, maximize the occurence in S of the distances between the N points.
For example:
S = {2, 4}, N = 4
One answer would be P = {2, 4, 6, 8}, since the distances between the points P are 2, 2, 2, 2, 4, 4, 6. Only 6 is not in S.
or
S = {7, 13, 14, 22}
N = 4501
answer ???
I'm not looking for an exact answer (although an exact answer wouldn't hurt) but rather I am trying to avoid reinventing the wheel (fun though it may be). What mathematical tools could I use to avoid bruteforcing the possible values of P. How should how would you approach this problem?
| https://mathoverflow.net/users/1214 | Given N points on a number line and m total distances between those points, are there efficent ways to optimize for particular values in m? | Some comments expanding on gowers' hunch: we may as well assume that the elements of S have gcd 1. If the size of the set is |S| = m and the sum of the elements of S is k then taking a block of N consecutive integers for $N \geq m$ gives (mN - k) pairs. There's also a theoretical maximum of $mN - \binom{m + 1}{2}$: every element of any N-set can be the larger half of at most m pairs, except that the smallest element can be the larger half of at most 0, the second-smallest can be the larger half of at most 1, etc., and the m-th smallest can be the larger half of at most (m - 1).
This theoretical maximum is achieved for S = {1, 2, ..., m} and matches gowers' strategy in this case. However, usually there's some gap between gowers' strategy and the theoretical max, and in some cases we can definitely do better than gowers' strategy: for example, with S = {1, 5, 10} and N = 12, a block of 12 consecutive integers gives us 11 + 7 + 2 = 20 pairs while an arithmetic progression with step size 5 gets us 11 + 10 = 21 pairs. So for small values of N we can sometimes do better by ignoring certain elements of S.
| 1 | https://mathoverflow.net/users/4658 | 18600 | 12,393 |
https://mathoverflow.net/questions/18588 | 18 | Then one can construct a model for the inverse limit by taking all the compatible sequences.
This is a subspace of a product of compact spaces. This product is compact by Tychonoff. If all the spaces are Hausdorff, then this is even a closed subspace.
However, if the spaces are not Hausdorff, it needn't be a closed subspace. If you take a two point space with the trivial topology as $X\_n$ and constant structure maps, you will get as the inverse limit the space of all constant sequences, which is not a closed subspace of the infinite product, as the infinite product also has the trivial topology.
But the space is again compact. So I am wondering, whether there is a generalization of the proof of Tychonoff's theorem, that applies directly to inverse limits.
| https://mathoverflow.net/users/3969 | Is a inverse limit of compact spaces again compact ? | What does this example do ...
All spaces are on set $\{1,2,\dots\}$. Space $X\_n$ has topology that makes $\{1,2,\dots,n\}$ discrete and $\{n+1,\dots\}$ indiscrete. Of course $X\_n$ is compact non-Hausdorff. Map $X\_{n+1} \to X\_n$ by the "identity". Inverse limit is ... ???
| 34 | https://mathoverflow.net/users/454 | 18605 | 12,397 |
https://mathoverflow.net/questions/18608 | 7 | We would like to know how hard it is to count Eulerian orientation in an undirected 4-regular graph. For a given edge orientation to be Eulerian, we mean that every vertex has 2 in-edges and 2 out-edges.
It is known that counting Eulerian orientation in undirected graphs are #P-complete. We have tried to construct some gadget to reduce the general case to 4-regular case, but did not succeed. Any idea about that?
Thank you.
| https://mathoverflow.net/users/4547 | Counting Eulerian Orientation in a 4-regular undirected graph | Let $G$ be a planar graph. Consider a [medial graph](http://en.wikipedia.org/wiki/Knots_and_graphs#Medial_graph) $H=H(G)$, which is always $4$-regular. Often, problems about $G$ can be translated into the language of $H$ and vice versa. Closer to your question, the number of Eulerian orientations of $H$ is "almost" an evaluation of the Tutte polynomial:
$$(\ast) \qquad \sum\_{O} 2^{\alpha(O)} = 2\cdot T\_G(3,3),$$
where the summation is over all Eulerian orientations $O$ of $H$, and $\alpha(O)$ is the number of *saddle vertices* (i.e. where the orientation is in-out-in-out in cyclic order). This is due to Las Vergnas (JCTB 45, 1988). My former student Mike Korn and I generalized this [here](http://www.math.ucla.edu/~pak/papers/tutte7color.pdf).
Of course, evaluations of the Tutte polynomial of planar graphs, including at ($3,3)$, are pretty much all #P-hard (with a few known exceptions), see D.J.A. Welsh, *Complexity: knots, colourings and counting* book (1993). Now, there is a bijective proof of $(\ast)$, which maps orientations $O$ into certain subsets of edges of $G$. It is possible that when you map the number of orientations without weight you still get a hard-to-compute stat. sum, which will prove what you want.
| 10 | https://mathoverflow.net/users/4040 | 18621 | 12,410 |
https://mathoverflow.net/questions/18568 | 12 | Hello,
I am investigating the Leech lattice. Lately I have discovered following. Some lattices decompose into distinct set of orthonormal frames. For example E8 lattice which contains 240 unitary vectors in dimension 8 decompose into 15 sets of 16 vectors in each set. Each set contain 16 vectors of +- orthonormal basis of R^8.
The numbers are:
* 24 = 3 \* 8, lattice in four dimension call it d4 lattice with vectors e1..e4, 1/2\* \*Sum(+-ei), i=1..4; it is root system of Lie algebra D4.
* 240 = 15\*16, E8 lattice
* 196560 = 4095\*48; Leech lattice.
My question is whether anybody knows similar decomposition of Leech lattice. I am trying to obtain one but no luck so far. Maybe it is already known fact.
Obviously each element of Conway group Co0 transform one orthonormal frame of Leech into another. So if I know the matrix representation of Co0 then I know many examples of such frames. Each element of Co0 would define permutation on 4095 points i.e. sets of orthonormal frames, so we would have homomorphism from Co0 to S4095.
Regards,
Marek Mitros
mim\_ (at) op.pl
| https://mathoverflow.net/users/4714 | Leech lattice decomposition | Yes, such a decomposition exists. Here's a construction I learned from Elkies some time ago (it's mentioned in one of his papers, probably *Mordell-Weil lattices in characteristic 2, II*), using an action of the Gaussian integers Z[i] on the Leech lattice:
Let L be the Leech lattice, and consider the quotient L/(1+i)L, which has order 2^12 (since 1+i has norm 2). No minimal vector in L can be in (1+i)L (the minimal vectors in L have norm 4, so those in (1+i)L have norm 8). Thus, the minimal vectors fall into at most 2^12 - 1 classes mod (1+i)L. Suppose v and w are minimal vectors that are congruent mod (1+i)L. Then |v-w|^2 >= 8, so the inner product is at most 0. There can be at most 48 such vectors on a sphere in R^24, so each residue class mod (1+i)L contains at most 48 minimal vectors. However, 196560 = (2^12 - 1)\*48, so there must be exactly 48 in each. This gives a decomposition into 4095 orthogonal frames.
| 18 | https://mathoverflow.net/users/4720 | 18626 | 12,415 |
https://mathoverflow.net/questions/18558 | 8 | I thought that I read a paper making this claim a few months ago, but now I can't find it. If the answer is yes, is there a nice way to go from the presentation of the right-angled coxeter group to a presentation of its right-angled artin subgroup? Thanks.
| https://mathoverflow.net/users/4027 | does every right-angled coxeter group have a right-angled artin group as a subgroup of finite index? | As James points out, the paper of Davis and Januskiewicz proves the inverse. To see that the answer to your question is 'no', consider the right-angled Coxeter group whose nerve graph is a pentagon. That is, it's the group with presentation
$\langle a\_1,\ldots, a\_5 \mid a\_i^2=1, [a\_i,a\_{i+1}]=1\rangle$
where the indices are considered mod 5.
This group acts properly discontinuously and cocompactly on the hyperbolic plane, and it's not hard to see that it has a finite-index subgroup which is the fundamental group of a closed hyperbolic surface. Every finite-index subgroup of a right-angled Artin group is either free or contains a copy of $\mathbb{Z}^2$, but the fundamental group of a closed hyperbolic surface has no finite-index subgroups of this form.
| 13 | https://mathoverflow.net/users/1463 | 18630 | 12,419 |
https://mathoverflow.net/questions/18636 | 18 | This question is inspired from [here](https://mathoverflow.net/questions/18547/number-of-unique-determinants-for-an-nxn-0-1-matrix), where it was asked what possible determinants an $n \times n$ matrix with entries in {0,1} can have over $\mathbb{R}$.
My question is: how many such matrices have non-zero determinant?
If we instead view the matrix as over $\mathbb{F}\_2$ instead of $\mathbb{R}$, then the answer is
$(2^n-1)(2^n-2)(2^n-2^2) \dots (2^n-2^{n-1}).$
This formula generalizes to all finite fields $\mathbb{F}\_q$, which leads us to the more general question of how many $n \times n$ matrices with entries in { $0, \dots, q-1$ } have non-zero determinant over $\mathbb{R}$?
| https://mathoverflow.net/users/2233 | Number of invertible {0,1} real matrices? | See [Sloane, A046747](http://oeis.org/A046747) for the number of singular (0,1)-matrices. It doesn't seem like there's an exact formula, but it's conjectured that the probability that a random (0,1)-matrix is singular is asymptotic to $n^2/2^n$.
Over $F\_2$ the probability that a random matrix is nonsingular, as $n \to \infty$, approaches the product $(1/2)(3/4)(7/8)\cdots = 0.2887880951$, and so the probability that a random large matrix is singular is only around 71 percent. I should note that a matrix is singular over $F\_2$ if its real determinant is even, so this tells us that determinants of 0-1 matrices are more likely to be even than odd.
| 22 | https://mathoverflow.net/users/143 | 18639 | 12,425 |
https://mathoverflow.net/questions/15282 | 18 | This is probably a very elementary question in symplectic geometry, a subject I've picked up by osmosis rather than ever really learning.
Suppose I have a symplectic manifold $M$. I believe that a *Lagrangian fibration* of $M$ is a collection of immersed Lagrangian submanifolds so that as a fibered manifold locally $M$ looks like a product. I.e. I can find local coordinates so that the fibers are $\{\text{half the coordinates} = \text{constant}\}$.
Then, at least locally, I can think about the set of fibers as some sort of "space" $N$. My question is: to what extent can I think of $M$ as the cotangent bundle $T^\*N$?
Surely the answer is "to no extent whatsoever" globally: the set of fibers is probably not a space in any good way, and certainly not a manifold (see: irrational line in a torus). But what about locally? Then it's really two questions:
>
> **Question 1:** If I have a lagrangian fibration in $M$, can I find local coordinates $p\_i,q^j: M \to \mathbb R$ so that the symplectic form is $\omega = \sum\_i dp\_i \wedge dq^i$ and the fibers are of the form $\{ \vec q = \text{constant}\}$.
>
>
>
I thought the answer was obviously "yes", and maybe it is, but what I thought worked I can't make go through all the way.
Then the question is about how canonical this is, and that's not really about general Lagrangian fibrations at all:
>
> **Question 2:** What is a good description of the local symplectomorphisms $\mathbb R^{2n} \to \mathbb R^{2n}$ of the form $\tilde q = \tilde q(q)$ and $\tilde p = \tilde p(p,q)$?
>
>
>
The beginning of the answer is that it is a local symplectomorphism if $\sum\_i d\tilde p\_i \wedge d\tilde q^i = \sum\_i dp\_i \wedge dq^j$, but the left-hand-side is $\sum\_{i,j,k} \bigl(\frac{\partial \tilde p\_i}{\partial q^j}dq^j + \frac{\partial \tilde p\_i}{\partial p\_j}dp\_j \bigr) \wedge \bigl( \frac{\partial \tilde q^i}{\partial q^k}dq^k \bigr)$, so the two conditions are that $\sum\_i \frac{\partial \tilde p\_i}{\partial q^j} \frac{\partial \tilde q^i}{\partial q^k}dq^k$ is a symmetric matrix, and that $\frac{\partial \tilde p}{\partial p}$ is the (maybe transpose, depending on your convention) inverse matrix to $\frac{\partial \tilde q}{\partial q}$.
Anyway, I guess for completeness I'll also ask the global question:
>
> **Question 0:** What global conditions on $M$ and the fibration assure that there is a global symplectomorphism with $T^\*N$ for some $N$?
>
>
>
| https://mathoverflow.net/users/78 | To what extent can I think of a Lagrangian fibration in a symplectic manifold as T*N? | Your Question 1 is called Darboux theorem for fibrations (see: Arnold, V., Givental, A., Symplectic geometry, Dynamical Systems IV, Symplectic Geometry and its Applications (Arnold, V., Novikov, S., eds.), Encyclopaedia of Math. Sciences 4, Springer-Verlag, Berlin-New York, 1990.)
Here is how to construct suitable Darboux coordinates. Let $q\_i$ be local coordinates in the base of the fibration, we identify them with their pullbacks to the symplectic manifold. The functions $q\_i$ generate Hamiltonian vector fields $X\_{i}$ and these fields are tangent to the fibers (note that $X\_{i}$'s commute). Let $\varphi\_{i}(t)$ be the flow map generated by $X\_{i}$ for time $[0,t]$.
Now we choose (locally) a Lagrangian submanifold $L$ transversal to the fibration. The coordinates $q\_i$ give coordinates on $L$, so $(q\_1,...,q\_n)$ stands for a point on $L$. Here is a construction of a local symplectomorphism $$(p\_1,...,p\_n,q\_1,...,q\_n) \mapsto \varphi\_{n}(p\_n)\circ ...\circ \varphi\_{1}(p\_1)(q\_1,...,q\_n).$$ It is easy to check that it is indeed a fibered symplectomorfism sending the symplectic structure to the standard one.
| 14 | https://mathoverflow.net/users/2823 | 18645 | 12,430 |
https://mathoverflow.net/questions/18644 | 11 | This question is closely related to [this](https://mathoverflow.net/questions/13813/construction-of-the-stiefel-whitney-and-chern-classes) previous question.
Chern and Stiefel-Whitney classes can be defined on bundles over arbitrary base spaces. (In Hatcher's Vector Bundles notes, he uses the Leray-Hirsch Theorem, which appears to require paracompactness of the base space. The construction in Milnor-Stasheff works in general, as does the argument given by Charles Resk in answer to the above question. A posteriori, this actually shows that Hatcher's construction works in general too, since he really just needs $w\_1$ and $c\_1$ to be defined everywhere.)
The proof of uniqueness (as discussed in Milnor and Stasheff, or in Hatcher's Vector Bundles notes, or in the answers to the above question) relies on the splitting principle, and hence (it seems to me) requires the existence of a metric on the bundle in question. More precisely, if we have two sequences of characteristic classes satisfying the axioms for, say, Chern classes, and we want to check that they agree agree on some bundle $E\to B$, the method is to pull back $E$ along some map $f: B'\to B$ (with $f^\*$ injective on cohomology) so that $f^\*E$ splits as a sum of lines. Producing the splitting seems to require a metric on $E$ (or at least on $f^\*E$).
If $B$ is not paracompact, bundles over $B$ may not admit a metric (and may admit a classifying map into the universal bundle over the Grassmannian), so my question is:
Are Chern and/or Stiefel-Whitney classes unique for arbitrary bundles? If not, do $w\_1$ and $c\_1$ at least determine the higher-dimensional classes?
| https://mathoverflow.net/users/4042 | Uniqueness of Chern/Stiefel-Whitney Classes | I'm going to assume that your characteristic classes are supposed to live in the *singular cohomology* of the base space. Then to show your uniqueness result, it should be enough if you can produce, for any space $B$, a map $f:B'\to B$ such that $B'$ is paracompact, and $f$ induces an isomorphism in singular cohomology.
For any $B$, you can find a CW-complex $B'$ and a weak equivalence $f: B'\to B$, by one of Whitehead's many theorems. Weak equivalences always induce isomorphisms in *singular* cohomology, and CW-complexes are paracompact (I think!). Hatcher's topology textbook proves all of these, except possibly for the paracompactness claim (for which I can't find a reference yet).
If you're talking about Cech cohomology, then this proof won't work.
| 17 | https://mathoverflow.net/users/437 | 18648 | 12,432 |
https://mathoverflow.net/questions/18653 | 7 | There is a beautiful way to see that the congruence subgroup $\Gamma(2)$ is free on two generators: the action of $\Gamma(2)$ on $\mathbb{H}$ is free and properly discontinuous, and there is a modular function $\lambda$ with respect to $\Gamma(2)$ coming from Legendre normal form such that $\mathbb{H}/\Gamma(2) \xrightarrow{\lambda} \mathbb{C} - \{ 0, 1 \}$ is an isomorphism. ([Details.](http://qchu.wordpress.com/2010/03/12/fractional-linear-transformations-and-elliptic-curves/)) It follows that $\mathbb{H}$ is the universal cover of $\mathbb{C} - \{0, 1 \}$, hence that $\Gamma(2)$ is isomorphic to the fundamental group of $\mathbb{C} - \{0, 1 \}$.
However, the action of $\Gamma(1) \simeq PSL\_2(\mathbb{Z})$ on $\mathbb{H}$ is not properly discontinuous free; there are problems, which maybe I should call "ramification," at the points $i, e^{ \frac{\pi i}{3} }, e^{ \frac{2\pi i}{3} }$. This is supposed to be responsible for the fact that $\Gamma(1)$ is not free, but is instead the free product of a cyclic group of order $2$ and a cyclic group of order $3$, where the former somehow comes from the behavior at $i$ and the latter somehow comes from the behavior at the sixth roots of unity. That's what I've been told, anyway, but I don't know how the argument actually goes. What general context does it fit into? (Is "monodromy" a keyword here?)
| https://mathoverflow.net/users/290 | How does this geometric description of the structure of PSL(2, Z) actually work? | The key word here is "Bass-Serre Theory" -- using the action on the hyperbolic plane, you can easily cook up a nice action of $PSL\_2(\mathbb{Z})$ on a tree. This is all described nicely in Serre's book "Trees".
EDIT: Let me give a few more details. It turns out that a group $G$ splits as a free produce of two subgroups $G\_1$ and $G\_2$ if and only if $G$ acts on a tree $T$ (nicely, meaning that it doesn't flip any edges) with quotient a single edge $e$ (not a loop) such that the following holds. Let $e'$ be a lift of $e$ to $T$ and let $x$ and $y$ be the vertices of $e'$. Then the stabilizers of $x$ and $y$ are $G\_1$ and $G\_2$ and the stabilizer of $e'$ is trivial.
If you stare at the fundamental domain for the action of $PSL\_2(\mathbb{Z})$ on the upper half plane, then you will see an appropriate tree staring back at you. There is a picture of this in Serre's book.
EDIT 2: This point of view also explains why finite-index subgroups $\Gamma$ of $PSL\_2(\mathbb{Z})$ tend to be free. If you restrict the action on the tree $T$ to $\Gamma$, then unless $\Gamma$ contains some conjugate of the order 2 or order 3 elements stabilizing the vertices, then $\Gamma$ will act freely. This means that the quotient $T/\Gamma$ will have fundamental group $\Gamma$. Since $T/\Gamma$ is a graph, this implies that $\Gamma$ is free.
| 10 | https://mathoverflow.net/users/317 | 18655 | 12,437 |
https://mathoverflow.net/questions/18609 | 19 | One of the reasons why the classical theory of binary quadratic forms
is hardly known anymore is that it is roughly equivalent to the theory
of ideals in quadratic orders. There is a well known correspondence
which sends the $SL\_2({\mathbb Z})$-equivalence class of a form
$$ (A,B,C) = Ax^2 + Bxy + Cy^2 $$
with discriminant
$$ \Delta = B^2 - 4AC $$
to the equivalence class of the ideal
$$ \Big(A, \frac{B - \sqrt{\Delta}}2\Big). $$
One then checks that this gives a group isomorphism between the
primitive forms with discriminant $\Delta$ and the class group
of ideals coprime to the conductor of the order with discriminant $\Delta$.
If we go from forms with integral coefficients to forms with
coefficients in a polynomial ring $k[T]$ for some field $k$,
then everything transforms nicely as long as $k$ has
characteristic $\ne 2$. In characteristic $2$, there is a well
known theory of function fields (instead of $Y^2 = f(T)$,
consider equations $Y^2 + Y h(T) = f(T)$), but I have no idea
what the right objects on the forms side are. There should be
decent objects, since, after all, calculations in the Jacobian
of hyperelliptic curves are performed essentially by composition
and reduction of forms. But I don't thinkg that $Ax^2 + Bxy + Cy^2$
will work; for example, these guys all have square discriminant.
Question: which objects (forms or something else?) correspond to
ideals in quadratic function fields with characteristic $2$?
Added: Thank you for the (three at this point) excellent answers.
I am now almost convinced that binary forms over $F\_2[T]$ actually
do work; the reason why I thought they would not was not so much the
square discriminant but the fact that the action of SL$\_2$ fixed the
middle coefficient $B$ of $(A,B,C)$, so I didn't get much of a
reduction. But now I see that perhaps this is not so bad after all,
and that some pair $(B,?)$ will be the analog of the discriminant in
the characteristic 2 case.
I knew about Kneser's work on composition, but was convinced (by a
lengthy article of Towber in Adv. Math.) that classical Gauss composition
of forms over PID's, which I was interested in, is quite far away from
composition of quadratic spaces over general rings. Special thanks to
Torsten for showing how to go from quadratic spaces back to forms!
| https://mathoverflow.net/users/3503 | Binary Quadratic Forms in Characteristic 2 | Starting, generally, with a commutative ring $R$ and a rank $2$ projective
module $P$ given with a quadratic form $\varphi$ we can form its Clifford algebra
$C(P)$. Its even part $S:=C^+(P)$ is then a commutative $R$ algebra of rank $2$.
If (which we may assume locally) $P=Re\_1+Re\_2$ we have that $S$ has basis
$1,e\_1e\_2=:h$ and
$h^2=e\_1e\_2e\_1e\_2=e\_1(\langle e\_1,e\_2\rangle-e\_1e\_2)e\_2=\langle e\_1,e\_2\rangle h-e\_1^2e\_2^2=\langle e\_1,e\_2\rangle h-\varphi(e\_1)\varphi(e\_2)$,
where $\langle-,-\rangle$ is the bilinear form associated to $\varphi$, giving an explicit
quadratic algebra over $R$. Furthermore, $C^-(P)=P$ and it thus becomes an
$C^+(P)$-module, explicitly
$h\cdot e\_1=e\_1e\_2e\_1=\langle e\_1,e\_2\rangle e\_1-e\_1^2e\_2=\langle e\_1,e\_2\rangle e\_1-\varphi(e\_1)e\_2$. Furthermore,
putting $L:=S/R$ we get an isomorphism $\gamma\colon \Lambda^2\_RP \to L$ given by $u\land v \mapsto
\overline{uv}$ (note that $u^2=\varphi(u)$ which maps to zero in $L$). Note for future
use that, putting $[u,v]:=\gamma(u\land v)$, we also have $[[u,v]u,u]=\varphi(u)[u,v]$, where
the left hand side is well-defined as
$[([u,v]+r)u,u]=[[u,v]u,u]+r[u,u]=[[u,v]u,u]$. We have therefore constructed
from $(P,\varphi)$ a triple $(S,P,\gamma)$, where $S$ is a quadratic (i.e., projective of
rank $2$) $R$-algebra, $P$ an $S$-module projective of rank $2$ as $R$-module
and an isomorphism $\gamma\colon \Lambda^2\_RP \to S/R$ of $R$-modules. Conversely, given such a
triple $(S,P,\gamma)$ there is a unique quadratic form $\varphi$ on $P$ such that
$[[u,v]u,u]=\varphi(u)[u,v]$ (where again the left hand side is well-defined). It is
easily verified that these constructions are inverse to each other.
In case $R=\mathbb Z$ I hope this gives the usual association between forms and modules over orders in quadratic number fields (but I admit shamelessly that I haven't checked it). When $R=k[t]$ I again hope this gives what we want.
In terms of algebraic group schemes and torsors we have the following situation.
If the quadratic form is perfect, we have that $S$ is an \'etale covering and
hence corresponds to an element of $H^1(\mathrm{Spec}R,\mathbb Z/2)$. It is the
image under $O\_2 \to \mathbb Z/2$ of the torsor in $H^1(\mathrm{Spec}R,O\_2)$
corresponding to $\varphi$. As $\mathbb Z/2$ also acts as an automorphism group of
$SO\_2$ (by conjugation in $O\_2$) we can use the torsor in
$H^1(\mathrm{Spec}K,\mathbb Z/2)$ to twist $SO\_2$ to get $SO(\varphi)$, the connected
component of $O(\varphi)$. Torsors over this group corresponds to isomorphism classes
of pairs consisting of a rank $2$ quadratic $R$-forms $\phi$ and an $R$-isomorphism
$C^+(\phi)\simeq S$. We have an alternative description of $SO(\varphi)$: We can consider the
units $T:=S^\ast$ of $S$ as an algebraic group and have a (surjective)
norm map $T \to \mathbb G\_m$ and then $SO(\varphi)$ is the kernel of this norm map.
Using the exact sequence $1\to SO(\varphi)\to T\to\mathbb G\_m\to1$ we see that an $SO(\varphi)$-torsor
is given by a projective $S$-module $P$ of rank $1$ together with the
choice of an isomorphism $\Lambda^2\_RP\simeq R$.
[Added] Overcoming some of my laziness I did the calculation in the integral
case ($R=\mathbb Z$ although the only thing we use is that $2$ is invertible):
Start with the quadratic form $Cx^2+Bxy+Ay^2$ (the switch between $A$ and $C$ is
to make my definition come out the same as the formula of the question). We then
have $h^2=Bh-AC$ so that $h=\frac{B-\sqrt{\Delta}}{2}$. We have $h\cdot e\_1=Be\_1-Ce\_2$ and
$h\cdot e\_2=Ae\_1$. This is just an abstract module over $S$ but we can make it a
fractional ideal by mapping $e\_2$ to $A$, then $e\_1$ maps to $h$ so that the
fractional ideal is indeed $(A,\frac{B-\sqrt{\Delta}}{2})$.
[Added later] One could say that the association of the ideal $(A,h)$ to the
quadratic form $Cx^2+Bxy+Ay^2$ over the ring $R[h]/(h^2-Bh+AC)$ is an answer to
the question which workds in all classical cases ($R=\mathbb Z$ and
$R=k[T]$). The reason that this looks simpler than the traditional ($R=\mathbb
Z$) answer is that we let the presentation of the quadratic order depend on the
quadratic form itself. Usually we have fixed the quadratic order and want to
consider all forms with this fixed quadratic order as associated order. This
means that we should fix some normal form for the order and then express the
ideal in terms of this normal. When $R=\mathbb Z$ orders are in bijection with
their discriminants $\Delta$ which are actual integers (as the discriminant is
well-determined modulo squares of units). A normal form for the order is
$\mathbb Z+\mathbb Z\sqrt{\Delta}$ or $\mathbb Z+\mathbb Z(\sqrt{\Delta}+1)/2$ depending
on whether $\Delta$ is even or odd. A curious feature of the form
$(A,\frac{B-\sqrt{\Delta}}{2})$ of the ideal is that the distinction between the odd
and even case is not apparent. However, the crucial thing is that it expresses a
$\mathbb Z$-basis for the ideal in terms of the canonical form of the quadratic
order.
The case of $R=k[T]$ for $k$ a field of odd characteristic is somewhat deceptive
as the discriminant $\Delta=B^2-4AC$ is not quite an invariant of the quadratic order
as there are units different from $1$ that are squares (except when $k=\mathbb
Z/3$!). Hence the formula $(A,\frac{B-\sqrt{\Delta}}{2})$ is not quite of the same
nature as for the $\mathbb Z$ case as there is a very slight dependence of $\Delta$
on the form (and not just on the order). We could fix that by choosing coset
representatives for the squares as a subgroup of $k^\ast$ and then the formula for
the ideal would would take the form $(A,\frac{B-\lambda\sqrt{\Delta}}{2})$ where $\Delta$ now
has been normalised so as to have its top degree coefficient to be a coset
representative.
The case when $k$ has characteristic $2$ is more complicated. We get and order
of the form $k[T][h]/(h^2+gh+f)$ but the question of a normal form is trickier.
The discriminant of the order (in the sense of the discriminant of the trace
form) is equal to $g^2$ and as all elements of $k$ are squares we can normalise
things so that $g$ is monic. However, the order is not determined by its
discriminant. This can be seen already in the unramified case when $g=1$ when a
normal form for $f$ is that it contain no monomials of even degree and the
constant term is one of a chosen set of coset representives for the subgroup
$\{\lambda^2+\lambda\}$ of $k$. We can decide to rather arbitrarily fix a generator $H$ for
every order with $H^2=GH+F$ where one sensible first normalisation is that $G$
be monic (for which we have to assume that $k$ is perfect). Then we have that $h=H+a$, with $a\in k[T]$, and the ideal would be
$(A,H+a)$.
There is a particular (arguably canonical) choice of $H$: We assume $G$ is monic
and then can write uniquely $G=G\_1G\_2\cdots G\_n$ with $G\_i$ monic and
$G\_{i+1}|G\_i$. The normal form is then that $F$ have the form
$F=G\_1F\_1+G\_1^2G\_2F\_2+\cdots+G\_1^2G\_2^2\cdots G\_nF\_n+G^2F'$ where $\deg F\_i<\deg G\_i$ and
$F'$ contain no square monomials and its constant term belongs to a chosen set of
coset representatives for $\{\lambda^2+\lambda\}$ in $k$.
One further comment relating to the classical formulas. When passing from a
fractional ideal to a quadratic form one classically divides by the norm of the
ideal (as is done in KConrad's reply). This means that the constructed form is
primitive, i.e., the ideal generated by its values is the unit ideal. Hence if
one starts with a quadratic form, passes to the ideal and then back to a
quadratic form one does not end up with the same form if the form is not
primitive. Rather the end form is the "primitivisation" where one has divided
the form by a generator for its ideal of values. This of course only makes sense
if the base ring is a PID. Even for a general Dedekind ring if one wants to work
with primitive forms one has to accept quadratic forms that take values in
general rank $1$ modules (i.e., fractional ideals).
The approach above makes another choice. It deals only with $R$-valued forms but
accepts non-primitive ones. This would seem to lead to a contradiction as
the classical construction leads to a bijection between modules and primitive
forms and the above leads to a bijection between modules and arbitrary
forms. There is no contradiction however (phew!) as the above construction leads
to smaller orders than the classical one in the non-primitive case.
Classically what one really works with (when $R$ is a PID) are lattices in $L$
(the fraction field of $S$), where a lattice $M$ is a finitely generated
$R$-submodule of $K$ containing a $K$-basis ($K$ the fraction field of $R$) for
$L$. The order one associates to $M$ is the subring of $L$ stabilising $M$. When
the quadratic form $\varphi$ is non-primitive $C^+(P)$ is not equal to this
canonically associated order but is a proper suborder by it. Dividing by a
generator for the ideal of values gives us a primitive form for which $C^+(P)$
is equal to the canonical order.
Finally there is a particular miracle that occurs for lattices in quadratic
extensions (of the fraction field of a Dedekind ring $R$), $M$ as a module over its stabilisation order is
projective. This is why classes of primitive forms with fixed order are in
bijection with the class group of the order.
| 13 | https://mathoverflow.net/users/4008 | 18657 | 12,438 |
https://mathoverflow.net/questions/18597 | 6 | Let $s\_{\lambda}$ and $m\_{\lambda}$ be the Schur and monomial symmetric functions indexed by an integer partition $\lambda$ ($\ell(\lambda)$ is the number of parts of $\lambda$ and $m\_i(\lambda)$ is the multiplicity of part $i$). By the hook-content formula we have:
$$
s\_{\lambda}(1^n) = \prod\_{u\in \lambda} \frac{n+c(u)}{h(u)},
$$
where $c(u)$ and $h(u)$ are the content and hook length of the cell $u\in \lambda$.
Using $s\_{\lambda} = \sum\_{\mu} K\_{\lambda \mu} m\_{\mu}$ where $K\_{\lambda \mu}$ is the Kostka number, the number of semistandard Young tableaux of shape $\lambda$ and type $\mu$. Then we get $\sum\_{\mu} K\_{\lambda \mu} m\_{\mu}(1^n)=\prod\_{u\in \lambda} \frac{n+c(u)}{h(u)}$. This counts semistandard Young tableaux of shape $\lambda$ and any type.
Does the sum $\sum\_{\mu}K\_{\lambda,\mu}$ have a known formula for $\ell(\lambda)\geq 2$? This would be the number of semistandard Young tableaux of shape $\lambda$ with partition type.
| https://mathoverflow.net/users/771 | Is there a known formula for the number of SSYT of given shape with partition type? | Let $k(\lambda)=\sum\_\mu K\_{\lambda\mu}$. Then we have the generating function
$\sum\_\lambda k(\lambda)s\_\lambda = \prod\_{n\geq 1} (1-h\_n)^{-1}$.
| 9 | https://mathoverflow.net/users/2807 | 18658 | 12,439 |
https://mathoverflow.net/questions/18666 | 10 | Does there exist a continuous function $f:[0,1]\rightarrow [0,1]$ such that $f$ takes every value in $[0,1]$ an infinite number of times?
| https://mathoverflow.net/users/4619 | Continuous function from $[0,1]$ to $[0,1]$ | Yes. In fact, there exists such an $f$ taking every value *uncountably* many times.
Take a continuous surjection $g: [0, 1] \to [0, 1]^2$. (Such things exist: they're space-filling curves.) Then the composite $f$ of $g$ with first projection $[0, 1]^2 \to [0, 1]$ has the required property.
| 25 | https://mathoverflow.net/users/586 | 18668 | 12,441 |
https://mathoverflow.net/questions/17569 | 7 | The recent article found [here](http://arxiv.org/abs/1002.3622) revisits Thomason's proof that symmetric monoidal categories model all connective spectra, but stops short of showing that there is a full closed model structure on this category (as does, it seems, Thomason's original paper.) Is there such a thing?
My guess is some lifting similar to how the model structure on small categories is derived would work, but I'm not sure if there are any complications.
| https://mathoverflow.net/users/4466 | A Model Structure on Symmetric Monoidal Categories | One basic problem is that the category of symmetric monoidal categories isn't complete. Its completion, in a basic sense, is the category of multicategories, on which it seems reasonable to conjecture there is a model category structure whose homotopy category "is" the connective part of stable homotopy -- we hope to prove this soon. See Elmendorf and Mandell, "Permutative categories, multicategories, and algebraic K-theory", which just appeared in Algebraic and Geometric Topology.
| 9 | https://mathoverflow.net/users/4732 | 18673 | 12,446 |
https://mathoverflow.net/questions/18661 | 5 | I am interested at the moment in what groups can occur as the fundamental group of a 4-manifold (or more generally, a 4-dimensional CW complex) with prescribed conditions on the intersection form. I have what I am hoping is a basic homotopy theory question:
A (orientable) PD-$n$ group is a group $G$ such that the Eilenberg-Maclane space $K(G,1)$ admits "Poincare duality", i.e. there is an $n$-dimensional integer homology class in $K(G,1)$ (thought of as the "fundamental class") such that cap product with it yields an isomorphism between the corresponding cohomology and homology groups (like for closed oriented manifolds). This is more general than saying that $K(G,1)$ admits the structure of an orientable closed manifold of dimension $n$.
Let $G$ be a PD-3 group. Is there any reason why $G$ cannot be the fundamental group of an orientable PD4 complex $X$ with vanishing second homotopy group, $\pi\_2(X)=0$?
| https://mathoverflow.net/users/4731 | PD3 groups and PD4 complexes | $G=Z^3$ is such an example. It is $\pi\_1(T^3)$ hence a PD-3 group. If $X$ is a Poincare 4-complex with fund group $Z^3$, then the injective (by Hopf) map on cohomology $H^2(G)\to H^2(X)$ cannot be onto, because its image is lagrangian for the intersection form by naturality of cup products. Dually $H\_2(X)\to H\_2(G)$ is not injective, and so $\pi\_2(X)$ is not zero. Following this kind of idea is what math.GT/0307101 and math.GT/0608103 is based on.
| 5 | https://mathoverflow.net/users/3874 | 18680 | 12,452 |
https://mathoverflow.net/questions/18667 | 3 | We can define the (first) homology of a surface $S$ by working with graphs embedded in $S$. That is, we take any (oriented) graph which is 2-cell embedded in $S$, and take cycles modulo boundaries in the usual way. Here, I am talking about homology with coefficients in $\mathbb{Z}$. The group that we get is independent of the graph, so is indeed a topological invariant of the surface.
I work with *group-labelled graphs*, which are oriented graphs with their edges labelled from a finite abelian group $\Gamma$. Proceeding as above, group-labelled graphs allow us to define *group-labelled surfaces*. That is, let $G$ be a $\Gamma$-labelled graph 2-cell embedded in a surface $S$. If each face of the embedding has group-value zero (the labels of edges on the boundary of the face sum to zero), then this gives a well-defined map on homology. In fact, the embedding of $G$ in S induces a homomorphism from $H\_1(S)$ to $\Gamma$. So, we can forget about the $\Gamma$-labelled graph and just study this homomorphism.
**My question is:** how does this construction relate to taking homology with coefficients from $\Gamma$?
Someone once told me that what I am really doing is working with *cohomology* with coefficients in $\Gamma$, but I didn't really get this. Can someone please clarify?
| https://mathoverflow.net/users/2233 | Homology with Coefficients | Hi Tony.
This is not really a homology-question, the core of it is the fundamental group. The homomorphism you are using is used in the study of [Van Kampen diagrams](http://en.wikipedia.org/wiki/Van_Kampen_diagram). Consider a presentation $G=\langle A|R\rangle$. A Van Kampen diagram on $S$ is a labeled graph like you have defined it. The only difference is that in a Van Kampen diagram all labels are generators (or their inverses) $a^{\pm 1}$ and not arbitrary words (although you could define it in this general way without problems because of the Van Kampen lemma).
Then every path in this graph has a group word written on it and "reading the word along a path" is a homomorphism {Paths}$\to G$ with respect to composition of paths. It turns out, that this is compatible with homotopy of paths so this induces a homomorphism $\pi\_1(S,x\_0)\to G$.
This is the general version of your homomorphism: If your $G$ happens to be abelian, then this homomorphism factorizes through $\pi\_1(S,x\_0)^{ab}$ which is $H\_1(S)$ by the Hurewicz theorem.
This point of view clarifies some connections between the geometry of Van Kampen diagrams and group theoretic questions.
For example the Van Kampen lemma tells you that a group word is trivial if and only if there is a Van Kampen diagram on this disk with this word written on the boundary.
Another fact is this one: If there are no nontrivial "reduced" Van Kampen diagrams on the torus, then every two commuting elements of $G$ generate a cyclic subgroup (i.e. $xyx^{-1}y^{-1}=1$ has only the trivial solutions $x=a^k, y=a^m$ for some $a\in G$.). In a similar spirit one can prove: If there are no nontrivial reduced Van Kampen diagrams on the real projective plane, then there are no involutions in $G$ (i.e. $x^2=1$ has only the trivial solution $x=1$), and if there are no nontrivial reduced Van Kampen diagrams on Klein's bottle, then the only element that is conjugated to its own inverse is the identity (i.e. $yxy^{-1}=x^{-1}$ has only the trivial solution $x=1$).
This connection between geometry and group properties becomes less obscure, if one knows the fundamental groups of the disk (1), the torus ($\langle x,y | xyx^{-1}y^{-1}=1\rangle$), the real projective plane ($\langle x | x^2=1\rangle$) and Klein's bottle ($\langle x,y | yxy^{-1}=x^{-1}\rangle$).
| 4 | https://mathoverflow.net/users/3041 | 18684 | 12,455 |
https://mathoverflow.net/questions/18631 | 5 | What is the mirror manifold of the total space of the bundle $O(-1)\oplus O(-1)$ over $P^1$? I have tried to find the answer on the web but failed. Is there a good reference for this? Thanks.
| https://mathoverflow.net/users/2380 | Mirror of local Calabi-Yau | The physicists (see e.g. [this paper of Aganagic and Vafa](http://arxiv.org/abs/hep-th/0012041)) will write the mirror as a threefold $X$ which is an affine conic bundle over the holomorphic symplectic surface $\mathbb{C}^{\times}\times \mathbb{C}^{\times}$ with discriminant a Seiberg-Witten curve $\Sigma \subset \mathbb{C}^{\times}\times \mathbb{C}^{\times}$. In terms of the affine coordinates $(u,v)$ on $\mathbb{C}^{\times}\times \mathbb{C}^{\times}$, the curve $\Sigma$ is given by the equation
$$
\Sigma : \ u + v + a uv^{-1} + 1 = 0,
$$
and so $X$ is the hypersurface in $\mathbb{C}^{\times}\times \mathbb{C}^{\times} \times \mathbb{C}^2$ given by the equation
$$
X : \ xy = u + v + a uv^{-1} + 1.
$$
From geometric point of view it may be more natural to think of the mirror not as an affine conic fibration over a surface but as an affine fibration by two dimensional quadrics over a curve. The idea will be to start with the Landau-Ginzburg mirror of $\mathbb{P}^{1}$, which is $\mathbb{C}^{\times}$ equipped with the superpotential $w = s + as^{-1}$ and to consider a bundle of affine two dimensional quadrics on $\mathbb{C}^{\times}$ which degenerates along a smooth fiber of the superpotential, e.g. the fiber $w^{-1}(0)$. In this setting the mirror will be a hypersurface in $\mathbb{C}^{\times}\times \mathbb{C}^{3}$ given by the equation
$$
xy - z^2 = s + as^{-1}.
$$
Up to change of variables this is equivalent to the previous picture but it also makes sense in non-toric situations. Presumably one can obtain this way the mirror of a Calabi-Yau which is the total space of a rank two (semistable) vector bundle of canonical determinant on a curve of higher genus.
| 11 | https://mathoverflow.net/users/439 | 18711 | 12,474 |
https://mathoverflow.net/questions/18698 | 14 | Let $\Sigma$ be an oriented topological surface. For simplicity, assume that the genus of $\Sigma$ is at least $2$. There are a number of classical results on the homotopy types of various groups of self-maps of $\Sigma$:
1) Earle and Eells proved that the components of $\text{Diff}(\Sigma)$ are contractible.
2) Hamstrom proved that the components of $\text{Homeo}(\Sigma)$ are contractible.
3) Peter Scott proved that the components of $\text{Homeo}^{\text{PL}}(\Sigma)$ are contractible, where by $\text{Homeo}^{\text{PL}}(\Sigma)$ we mean the group of PL self-homeomorphisms of $\Sigma$.
Of course, in 1 and 3 we are fixing a $C^{\infty}$ or $\text{PL}$ structure on $\Sigma$, respectively.
Another standard fact is that every self homotopy-equivalence of $\Sigma$ is homotopic to a homeomorphism. This leads me to my question. Denote by $\text{HE}(\Sigma)$ the set of self homotopy-equivalences of $\Sigma$. Are the components of $\text{HE}(\Sigma)$ contractible?
Another related question is as follows. There is a beautiful alternate proof of the above theorem of Earle and Eells due to Earle and McMullen (see their paper "Quasiconformal Isotopies"; their proof uses complex analysis). The proofs of Hamstrom's theorem and Scott's theorem are very complicated -- are there any alternate approaches to them in the literature?
| https://mathoverflow.net/users/317 | Homotopy type of set of self homotopy-equivalences of a surface | A couple comments. For the result about diffeomorphism groups there is a very nice alternative proof due to A. Gramain in the Annales Scient. E.N.S. v.6 (1973), pp. 53-66, that uses no analysis, just basic differential topology. Another approach, which I'm not sure is written down anywhere in detail, is to take the proof of the corresponding result for Haken 3-manifolds (due independently to Ivanov and myself) and scale it back from 3-manifolds to surfaces. This too uses no analysis, just basic topology. I don't recall Scott's method for the PL case, but it might be similar to this.
For the result about homotopy equivalences it's best to look first at the space of homotopy equivalences that fix a basepoint. This has contractible components for any $K(\pi,1)$ space, by an elementary obstruction theory argument. (Just deform families of maps to the identity map, cell by cell, which is possible since the obstructions lie in the higher homotopy groups of the space.) By evaluating arbitrary homotopy equivalences at the basepoint one gets a fibration where the total space is the space of all homotopy equivalences, the base space is the $K(\pi,1)$ space, and the fiber is the earlier space of basepoint-preserving homotopy equivalences. Looking at the long exact sequence of homotopy groups for this fibration then gives the desired result. Triviality of the center of $\pi$ comes in at this point to show that the boundary map from $\pi\_1$ of the base to $\pi\_0$ of the fiber is injective.
| 15 | https://mathoverflow.net/users/23571 | 18715 | 12,478 |
https://mathoverflow.net/questions/18677 | 8 | Can anyone provide me with the reference for the following fact
(idea of the proof will be appreciated too):
Cohomology ring with $\mathbb Q$-coefficients of a group $G$ (I don't know precisely what the assumptions are: reductive complex algebraic group or maybe complex Lie group G with some restrictions. The cases I'm interested in are $GL\_n(\mathbb C)$ and $SL\_n(\mathbb C)$) is the exterior algebra on the generators of odd degrees, with the number of generators equal to the rank of $G$.
This fact is attributed to H.Hopf, but I wasn't able to find a reference.
Thanks.
| https://mathoverflow.net/users/2260 | Cohomology rings of $ GL_n(C)$, $SL_n(C)$ | If $G$ is a connected Lie group (or just a connected loop space with finite homology) then $H^\*(G,\mathbf{Q})$ is a Hopf algebra. Graded connected Hopf algebras over $\mathbf{Q}$ are always tensor products of exterior algebras in odd degrees with polynomial algebras in even degrees. Since polynomial algebras are infinite, they can't occur. The reference to Hopf is probably H. Hopf, Über die algebraische Anzahl von Fixpunkten, Math. Z. 29 (1929), 493–524. For the classification of Hopf algebras, see also A. Borel, Sur la cohomologie des espaces fibrés principaux et des espaces homogènes de
groupes de Lie compacts, Ann. of Math. (2) 57 (1953), 115–207.
| 11 | https://mathoverflow.net/users/4183 | 18721 | 12,481 |
https://mathoverflow.net/questions/17980 | 0 | From what I understand, an exotic n-sphere is a manifold which is homeomorphic to the n-sphere but not diffeomorphic to it.
Now I have read that there are no exotic 2-spheres.
But isn't something like a tetrahedron an example of a manifold which is homeomorphic to the sphere, but not diffeomorphic ? (Because of the corners and edges.)
What am I missing ?
| https://mathoverflow.net/users/4279 | Question about exotic spheres | A exotic sphere is (by definition) a differentiable manifold. So if you want to consider the tetrahedron, you have to specify, what differentiability at one of the edges means. As soon as you specified this, you will just get the 2-sphere.
| 5 | https://mathoverflow.net/users/3969 | 18724 | 12,482 |
https://mathoverflow.net/questions/18723 | 14 | Often, certain symbols in mathematics denote different things in different fields. Is there any sort of ordered list that will tell you what a certain symbol means in alphabetical order by the symbol's alias in LaTeX, perhaps with the way to pronounce it out loud?
I'm thinking of something like this [Wikipedia page](http://en.wikipedia.org/wiki/Table_of_mathematical_symbols) but more comprehensive and usefully ordered by LaTeX alias (The one on wikipedia has very few symbols, and I am familiar with all of them already). The problem is that when you want to find the meaning of a symbol, there is no way to search on google (because google has no support at all for searching for symbols). Oftentimes, I'm forced to ask someone around the department what it means or how to say it out loud.
For example, I'm trying to find the meaning of the symbol $\uplus$, but I have no way of finding out what it means. Also, for the longest time, I couldn't figure out what to call $f\_!$ or $f^!$. How should I know that they're called "f lower shriek" or "f upper shriek".
So for the question: Does any such list exist for either pronunciation, meaning, or both (aside from the one on Wikipedia that I just noted)?
| https://mathoverflow.net/users/1353 | Mathematical symbols, their pronunciations, and what they denote: Does a comprehensive ordered list exist? | Comments suggest this ...
<http://www.fileformat.info/info/unicode/category/Sm/list.htm>
However, a designation like "rightwards arrow above reverse tilde operator" doesn't really answer the question here, does it?
| 5 | https://mathoverflow.net/users/454 | 18731 | 12,488 |
https://mathoverflow.net/questions/18701 | 7 | Let $(M,g)$ be a Riemannian manifold, $p$ a point on the manifold and $v \in T\_p M$. Let $\gamma$ be the geodesic starting at $p$ in the direction $v$. There exists a time $t\_f$ such that there exists a [Fermi coordinate system](http://en.wikipedia.org/wiki/Fermi_coordinates) adapted to $\gamma$ up to time $t\_f$.
**My question:** Does there exist a lower bound for $t\_f$ in terms of the 2-jet of $g$ at $p$?
That is, I have solid estimates on $g$ up to its second derivatives at $p$: $$\|g\| + \|\nabla g\| + \|\nabla^2 g\| \le h \qquad (1)$$ for some $h$. I would like to show that there exists $f(h)$ such that $$t\_f \ge f(h)$$ for all Riemannian metrics satisfying (1) at $p$.
**Edit** (Mar 19): Taking the helpful advice of Anton, Deane, TK and Willie into account, I've reworded the question:
Let $U = B(0,r)$ be the closed Euclidean ball of radius $r$ in $\mathbb R^n$. Write $$\lambda = \inf\_{x \in U} \inf\_{\|v\|=1} \langle v,v \rangle\_{g(x)}$$ as the minimum eigenvalue of the metric in $U$, and suppose that $$\frac{1}{\lambda} + \|g\|\_{C^2(U)} \le h.$$ This is a more refined version of (1) above. Since $$\ddot \gamma^k = -\Gamma^k\_{ij} \dot \gamma^i \dot \gamma^j,$$ our estimate gives a control on the acceleration of a geodesic $\gamma$ in $U$, so there exists a minimum self-intersection time $t\_i$ depending on $h$ and $r$ (i.e., if $t, t' \le t\_i$ then $\gamma(t) \ne \gamma(t')$).
Does this imply the existence of a uniform lower bound on $t\_f$ (depending only on $r$ and $h$)? If so, can we relax the control on the second derivative of $g$?
More succinctly: Are existence and non-self-intersection of a geodesic the only obstructions to the existence of Fermi coordinates?
| https://mathoverflow.net/users/238 | Existence of Fermi coordinates on a Riemannian manifold | I would answer this question this way:
First, without any further assumptions about the Riemannian metric on $M$, you don't even have a lower bound on the time $t\_f$ for which the geodesic *exists* and does not intersect itself. The geodesic may fail to exist either simply because the metric is incomplete.
Second, if for some reason you know that the geodesic exists for time $t\_f$ and does not hit any boundary or singular set of $M$, then you can definitely find Fermi co-ordinates on a neighborhood of the geodesic, *but* you have no control over the "thickness" of the neighborhood along the geodesic, so it may approach zero.
Third, the answers above hold regardless of whether you know anything about the metric at a single point $p$. That is far too little information, and you can make the metric do anything you want at a point arbitrarily close to the point $p$. As others have indicated, you must assume some information on an neighborhood of $p$, and the conclusion holds only for that neighborhood.
ADDED: Your question is analogous to the following one: Suppose I have a real-valued function $f$ on an interval, and for some (large) value of $k$, I know the $k$-th order Taylor polynomial of $f$ at a point $p$ in the interval. Is there a bound on the first derivative of $f$ on an open interval containing $p$, where the bound and the size of the interval depend only on the coefficients of the Taylor polynomial?
It is easy to show that the answer is no, and essentially the same proof can be used to show that the answer to your question is also no.
RESPONSE TO REVISED VERSION:
1) Yes, if your manifold is open, and you start with a geodesic segment, then there exist Fermi co-ordinates in a neighborhood of the geodesic. But you have no control over the thickness of the neighborhood around the geodesic on which the Fermi co-ordinates exist. It will vary and may approach zero as you approach one of the endpoints of the geodesic segment.
2) I'm a little baffled by why you would want to do everything relative to a background flat metric. Is this metric a natural part of what you need this for? This makes the question a bit contrived for me.
| 6 | https://mathoverflow.net/users/613 | 18737 | 12,492 |
https://mathoverflow.net/questions/18734 | 3 | I am trying to produce closed quotient maps, as they allow a good way of creating saturated open sets (as in this [question](https://mathoverflow.net/questions/17612/finding-saturated-open-sets)).
A map $f:X\rightarrow Y$is called proper, iff preimages of compact sets are compact. It is called quotient map, iff a subset $V\subset Y$ is open, if and only if its preimage $f^{-1}(V)$ is open. And it is called closed, iff it maps closed sets to closed sets.
So the question is, whether a proper quotient map is already closed.
Note that, I am particular interested in the world of non-Hausdorff spaces.
| https://mathoverflow.net/users/3969 | Is a proper quotient map closed ? | No. Let X={1,2,3} and Y={1,2}. Let f map 1 to 1, 2 and 3 to 2. Let the topology on X be {∅,{2},{1,2},{2,3},{1,2,3}} and that on Y be {∅,{2},{1,2}}. f maps the closed set {3} onto the non-closed set {2}.
| 3 | https://mathoverflow.net/users/802 | 18742 | 12,496 |
https://mathoverflow.net/questions/18719 | 3 | Let $ K $ be a finite extension of a $p$-adic field or a number field, L a finite extension of $K$. The following fact holds: $
\text{Gal}(K^{\text{ab}} / K) \rightarrow \text{Gal}(L^{\text{ab}} / L)
$, where the arrow is the Transfer (Verlagerung) map, is injective.
I wonder whether this is an arithmetic fact or a fact about absolute Galois group of fields in general?
| https://mathoverflow.net/users/2701 | Injectivity of Transfer (Verlagerung) map | I'm not sure how to answer the more philosophical question (it's likely you could encode enough of the axioms to force the purely group-theoretic version of the question to be true, but to ask whether that's what's "really" going on....), but it's certainly not true for all pairs of groups that fit in a similar commutative diagram -- in fact, it's not even true for all such pairs of groups coming from similar questions in algebraic number theory. For example, instead of taking the maximal abelian extension of $K$, take the maximal abelian extension of $K$ which is unramified outside of a set of primes, or split completely at a set of primes, or both -- and you'll pick up a kernel to you transfer map (see Gras, Class Field Theory for some specific calculations of kernels like this). A very relevant related topic worth bringing up is the theorem of Gruenberg-Weiss, which gives an impressively vast generalization of the group-theoretic (and hence the ideal-theoretic) principal ideal theorem entirely in terms of kernels of related transfer maps.
| 5 | https://mathoverflow.net/users/35575 | 18745 | 12,498 |
https://mathoverflow.net/questions/18748 | 8 | Let's say we want to define a choice function for certain particular subsets $S \subset2^{\mathbb{R}}$, i.e. we want a function $c:S \rightarrow \mathbb{R}$ such that $c(X)\in X$ for every $X\in S$. We don't want to invoke the axiom of choice. Clearly we require $\emptyset\notin S$.
For example, if $S$ is the set of non-empty open sets for the usual topology, then we can fix an enumeration of the rationals and for every open $A$ pick the first rational (in this particular enumeration) lying in $A$.
If $S$ is the set of non-empty closed sets, then for any $A\in S$ we can consider the least $n$ such that $\[-n,n\]\cap A$ is non empty and then pick the infimum of this non empty compact set.
The question is the following: can you define a choice function for, say, $S=F\_{\sigma}\setminus \{\emptyset\}$ or $G\_{\delta}\setminus\{\emptyset\}$ or maybe for the higher levels of the Borel hierarchy? Is it possible to prove that such choice function exists for such $S$ without using the axiom of choice?
| https://mathoverflow.net/users/1049 | Choice function for Borel sets? | It is impossible without using the Axiom of Choice to prove the existence of such a choice function for $F\_{\sigma}$
sets, which include the countable sets. To see this, it is easier to work with the space $2^{\mathbb{N}}$ of infinite binary sequences instead of $\mathbb{R}$. Let $E$ be the eventual equality equivalence relation on $2^{\mathbb{N}}$. Then there is no measurable function which selects an element from each $E$-class.
| 6 | https://mathoverflow.net/users/4706 | 18755 | 12,501 |
https://mathoverflow.net/questions/18747 | 15 | Well, I don't have any notion of *arithmetic geometry*, but I would like to understand what arithmetic geometers mean when they say "integer point of a variety/scheme $X$" (like e.g. in "integer points of an elliptic curve").
Is an integer point just defined as a morphism from $Spec\mathbb{Z}$ into $X$?
Suppose $X$ is given a structure of variety over, say, $Spec\mathbb{C}$; does the notion of integer point interact with this structure?
How is all of this related to finding the integer solutions of a concrete polynomial equation (perhaps with integer coefficients)?
Now a very, *very*, naif question. Suppose you have the plane $\mathbb{A}^{2}\_{\mathbb{C}}$ and draw two lines on it: $X=\{x=0\}$ and $X'=\{x=\pi\}$ (with the obvious reduced induced closed subscheme structure). Well, $X$ and $X'$ are clearly isomorphic as schemes over $\mathbb{C}$ (hence as schemes). But, if having integer points is somehow related to finding integer solutions to equations, how do you explain that $X$ (as embedded in $\mathbb{A}^{2}$) has plenty of points with integer *coordinates*, while $X'$ has none?
This to mean: how can the intrinsic notion of a morphism from $Spec\mathbb{Z}$ be possibly related to finding solutions to concrete (coordinate dependent) equations?
| https://mathoverflow.net/users/4721 | Integer points (very naive question) | To make sense of the notion of integer points, your scheme should be defined over $\mathbb{Z}$. What do we mean by that? Of course we should not ask for a structure map tp $Spec(\mathbb{Z})$, since every scheme has one such map. The right notion is the following.
Let $X$ be a scheme over $\mathbb{C}$; so by definition we have a structure map $X \to \mathop{Spec} \mathbb{C}$. Then we say that $X$ is defined over $\mathbb{Z}$ is there exists a scheme $X\_{\mathbb{Z}}$ over $\mathbb{Z}$ such that $X$ is the base change of $X\_{\mathbb{Z}}$ to $\mathbb{C}$, i. e. $X \cong X\_{\mathbb{Z}} \times\_{\mathbb{Z}} \mathop{Spec} \mathbb{C}$.
Now for such a scheme an integral point is a map $\mathop{Spec}\mathbb{Z} \to X\_{\mathbb{Z}}$ such that the composition with the structure map is the identity. Note that the same can be done for every ring $A$ in place of $\mathbb{Z}$.
With this definition, the line $\{ x = 0 \}$ is defined over $\mathbb{Z}$, but the line $\{ x = \pi \}$ is not, basically because there is no way to generate its ideal with equations having integer coefficients. So your problem does not arise anymore.
EDIT: Abstractly of course the two lines are isomorphic over $\mathbb{C}$, so the line $r = \{x = \pi \}$ actually has a model over $\mathbb{Z}$. The problem is that this model is not compatible with the inclusion in $\mathbb{A}^2$, that is, there will be no map $r\_{\mathbb{Z}} \to \mathbb{A}^2\_\mathbb{Z}$ whose base change is the inclusion of $r$ into $\mathbb{A}^2$. In order to have this, you would have to ask that the ideal of $r$ in $\mathbb{A}^2$ should be generated by polynomials with integer coefficients.
As for your second question, there can be different models, that is, nonisomorphic schemes over $\mathbb{Z}$ which become isomorphic after base change to $\mathbb{C}$. So before discussing the existence of integral points, you have to FIX a model, and the points will in general depend on the model.
For instance take the two conics $\{ x^2 + y^2 = 2 \}$ and $\{ x^2 + y^2 = 3 \}$. Both have an obvious choice of a model, given by the inclusion in $\mathbb{A}^2$; moreover they are isomorphic over $\mathbb{C}$. But the integral points on the first one are $(\pm 1, \pm 1)$, while the second has none.
Finally you consider the possibility that the structure over $\mathbb{C}$ is not relevant. This is false: the base change $X\_\mathbb{Z} \times\_{\mathbb{Z}} \mathop{Spec} \mathbb{C}$ is endowed with a natural map to $\mathop{Spec} \mathbb{C}$, and we ask for the isomorphism with $X$ to be over $\mathbb{C}$.
| 22 | https://mathoverflow.net/users/828 | 18757 | 12,502 |
https://mathoverflow.net/questions/18756 | 0 | Consider the [2D Shubert function](http://www.staff.brad.ac.uk/jpli/research/scga/shubert/shubrt21.htm). As given in that page, the function has 18 global minima and several local minima. How can I find the (x,y) of all these minima? Any help appreciated. If it was a summation (instead of a product), I would have done it by minimizing each individual term. However, I have 0 clue as to how to find the minima in this case.
UPDATE: Before applying any global optimizer, I want to know "theoretically" what are the (x,y) of all the minima. I want to be able to compare the expected and the obtained results
| https://mathoverflow.net/users/3552 | Finding the local/global minima of Shubert function | At Jacques' cajoling, I'm turning the comments into an answer.
The two dimensional Shubert function is just the product of the one dimensional one by itself. $f(x,y) = g(x)g(y)$ where $g(x) = \sum\_{j = 1}^5 j \cos( (j+1)x + j)$ is the 1 dimensional Shubert function. Observe that the local maxima are all positive, and the local minima all negative. So the minima for $f(x,y)$ occur at points $\{(x,y) : g'(x)= g'(y) = 0, f(x,y) < 0\}$. In other words, the minima of $f$ occurs at points where a maximum of $g$ is multiplied against a minimum.
Notice that there are 3 global max/min each of $g$ in the interval (-10,10), and 19 max and 20 min overall. This produces the 760 total local min of $f$ with 18 of them being global. (760 = 2 \* 19 \* 20, 18 = 2 \* 3 \* 3)
To find the extrema of the 1-d Shubert function, you evaluate its first derivative, and find that it can be simplified to a degree 6 polynomial in $\sin(x)$ and $\cos(x)$ by using the angle addition formulae. I have not evaluated the computations myself, so cannot tell you whether the expression has a closed-form algebraic solution.
| 2 | https://mathoverflow.net/users/3948 | 18762 | 12,505 |
https://mathoverflow.net/questions/18772 | 12 | Suppose $G$ is a semisimple group, and $V\_{\lambda}$ is an irreducible finite-dimensional representation of highest weight $\lambda$. Suppose $H \subset G$ is a semisimple subgroup. What is the multiplicity of the trivial representation in $V\_{\lambda}|\_{H}$? Is there a simple way to read this off from $\lambda$ and the Dynkin diagrams of $G$ and $H$?
| https://mathoverflow.net/users/1464 | Occurrence of the trivial representation in restrictions of Lie group representations | The short answer is: no. In theory, there are explicit formulae for branchings based on the Weyl character formula, but no reasonable person would call these simple.
There are interesting results which give you some information about branching multiplicities, but all involve real work. For instance, if G and H are compact, you can obtain asymptotic information about H-invariants in the representations $V\_{n\lambda}$ as a function of $n$: it's a polynomial, whose leading order is the dimension of $\mathcal{O}\_{\lambda}//H $, the symplectic reduction of the coadjoint orbit through $\lambda$ by $H$, and whose leading coefficient is the symplectic volume of this manifold.
If you prefer algebraic geometry, this polynomial is the Hilbert polynomial of the corresponding GIT quotient.
If $H$ is a root subalgebra, then life is a bit easier, and you can use combinatorial methods like crystals, but this is still not easy.
| 11 | https://mathoverflow.net/users/66 | 18779 | 12,513 |
https://mathoverflow.net/questions/18774 | 6 | Is there some **onto** function $f:$ $\mathbb{C}$ $\rightarrow$ $\mathbb{C}$
such that for each triangle $T$ (with its interior), $f(T)$ is a
square (with interior, too) ?
I would have the same question for triangles and squares *without* interior, respectively.
| https://mathoverflow.net/users/4750 | Triangles, squares, and discontinuous complex functions | With interior: yes. Fix a sequence of squares $Q\_1\subset Q\_2\subset\dots$ whose union is the entire plane. Then arrange a map $g:\mathbb R\to\mathbb R^2$ such that, for every nontrivial segment $[a,b]\subset\mathbb R$, its image is one of the squares $Q\_i$. To do that, construct countably many disjoint Cantor sets so that every nontrivial interval contains at least one of them. Then send every Cantor set $K$ bijectively onto $Q\_n$ where $n$ is the minimum number such that $K\cap [-n,n]\ne\emptyset$. Send the complements of these Cantor sets to a fixed point inside $Q\_1$. Then define $f(x,y)=g(y)$.
(This is a detailed version of gowers' answer.)
UPDATE
Without interior: no. Take any triangle $T$ and consider its image $Q$ with vertices $ABCD$. There is a side $I$ of $T$ whose image has infinitely many points on (at least) two sides of $Q$. If these are opposite sides, say $AB$ and $CD$, the image of any triangle containing $I$ must stay within the strip bounded by the lines $AB$ and $CD$. And if these are two adjacent sides of $Q$, say $AB$ and $AD$, the image of any triangle containing $I$ stays within the quarter of the plane bounded by the rays $AB$ and $AD$. In both cases, the images of the triangles containing $I$ do not cover the plane, hence the map is not onto.
| 10 | https://mathoverflow.net/users/4354 | 18783 | 12,515 |
https://mathoverflow.net/questions/18716 | 20 | I had been looking lately at Sylow subgroups of some specific groups and it got me to wondering about *why* Sylow subgroups exist. I'm very familiar with the proof of the theorems (something that everyone learns at the beginning of their abstract algebra course) -- incidentally my favorite proof is the one by Wielandt -- but the statement of the three Sylow theorems still seems somewhat miraculous. What got Sylow to imagine that they were true (especially the first -- the existence of a sylow subgroup)? Even the simpler case of Cauchy's theorem about the existence of an element of order $p$ in a finite subgroup whose order is a multiple of $p$ although easy to prove (with the right trick) also seems a bit amazing. I believe that sometimes the hardest part of a proving a theorem is believing that it might be true. So what can buttress the belief for the existence of Sylow subgroups?
| https://mathoverflow.net/users/2784 | Sylow Subgroups | Victor, you should check out Sylow's paper. It's in Math. Annalen 5 (1872), 584--594. I am looking at it as I write this. He states Cauchy's theorem in the first sentence and then says "This important theorem is contained in another more general theorem: if the order is divisible by a prime power then the group contains a subgroup of that size." (In particular, notice Sylow's literal first theorem is more general than the traditional formulation.) Thus he was perhaps in part inspired by knowledge of Cauchy's theorem.
Sylow also includes in his paper a theorem of Mathieu on transitive groups acting on sets of prime-power order (see p. 590), which is given a new proof by the work in this paper. Theorems like Mathieu's may have led him to investigate subgroups of prime-power order in a general finite group (of substitutions).
| 20 | https://mathoverflow.net/users/3272 | 18788 | 12,517 |
https://mathoverflow.net/questions/18787 | 21 | Here's a question I can't answer by myself: The Reflection Principle in Set Theory states for each formula $\phi(v\_{1},...,v\_{n})$ and for each set M there exists a set N which extends M such that the following holds
$\phi^{N} (x\_{1},...,x\_{n})$ iff $\phi (x\_{1},...,x\_{n})$ for all $x\_{1},...x\_{n} \in N$
Thus if $\sigma$ is a true sentence then the RFP yields a model of it and as a consequence any finite set of axioms of ZFC has a model (as a consequence ZFC is not finitely axiomatizable by Gödel's Second Incompleteness Theorem)
But why can't I just use now the Compactness Theorem (stating that each infinte set of formulas such that each finite subset has a model, has a model itself) to obtain a model of ZFC (which is actually impossible)??
| https://mathoverflow.net/users/4753 | Montague's Reflection Principle and Compactness Theorem | For any finite set of axioms K of ZFC, ZFC proves "K has a model", via the reflection principle as you note. However, ZFC does not prove "for any finite set of axioms K of ZFC, K has a model". The distinction between these two is what prevents ZFC from proving that ZFC has a model.
(That is, even though, as you note, ZFC proves "if every finite set of axioms K of ZFC has a model, then ZFC has a model", as ZFC proves compactness, it does not follow that ZFC proves the consequent of this implication, as in fact ZFC does not prove the antecedent; ZFC only proves each particular instance of the antecedent, but not the universal statement itself.)
| 28 | https://mathoverflow.net/users/3902 | 18793 | 12,520 |
https://mathoverflow.net/questions/18753 | 14 | Consider a compact manifold *M*. For a vector field *X* on *M*, let $\phi\_X$ denote the diffeomorphism of *M* given by the time 1 flow of *X*.
If *X* and *Y* are two vector fields, is $\phi\_X \circ \phi\_Y$ necessarily of the form $\phi\_Z$ for some vector field *Z*?
Since $X\mapsto \phi\_X$ can be thought of as the exponential map from the Lie algebra of vector fields to the group of diffeomorphisms, an obvious candidate is that *Z* should be given by the Baker-Campbell-Hausdorff formula $B(X, Y) = X+Y+\frac{1}{2}[X,Y]+\cdots$. But does this hold in this infinite-dimensional setting? If so, in which sense does the series converge to *Z*?
Also, I'm interested in the case where *M* is a symplectic manifold and we consider only symplectic vector fields (ie. vector fields for which the contraction with the symplectic form is a closed 1-form). Locally, *X* and *Y* are the Hamiltonian vector fields associated to smooth functions *f* and *g*, so I assume that asking whether *B(X, Y)* makes sense/is symplectic corresponds to asking whether *B(f, g)* makes sense/defines a smooth function (where, of course, we use the Poisson bracket in the expansion of *B(f, g)*). The right-hand side of *B(f,g)* consists of lots of iterated directional derivatives of *f* and *g* in the *Xf* and *Xg* directions; it is not clear to me that the coefficients in the BCH formula make the series converge (uniformly, say) for any choice of *f* and *g*.
| https://mathoverflow.net/users/4747 | Does the Baker-Campbell-Hausdorff formula hold for vector fields on a (compact) manifold? | To answer your first question, the composition of two time-1 flows won't necessarily be another time-1 flow.
One way to see this is to note that when a time-1 flow $\phi\_X$ has a periodic point $P$ (period > 1), then $P$ can't be hyperbolic since it lies on a closed orbit of the flow for $X$. (The eigenvector of $D\phi\_X$ tangent to this orbit has corresponding eigenvalue 1.)
Now, take a flow on $S^2$ whose time-1 map rotates the sphere, switching the north and south poles. Take a second flow for which both poles are hyperbolic attracting fixed points. Composing the two time-1 maps gives you a new diffeomorphism with hyperbolic points of period 2.
| 12 | https://mathoverflow.net/users/1227 | 18801 | 12,524 |
https://mathoverflow.net/questions/18799 | 39 | The push-pull formula appears in several different incarnations. There are, at least, the following:
1) If $f \colon X \to Y$ is a continous map, then for sheaves $\mathcal{F}$ on $X$ and $\mathcal{G}$ on $Y$ we have $f\_{\*} (\mathcal{F} \otimes f^{\*} \mathcal{G}) \cong f\_{\*} (\mathcal{F}) \otimes \mathcal{G}$.
A similar formula holds for the derived functors and for $f^{!}$.
2) If $f \colon X \to Y$ is a proper map of schemes, with $Y$ smooth, both $f^{\*}$ and $f\_{\*}$ are defined on the Chow groups, and $f\_{\*}(\alpha \cdot f^{\*} \beta) = f\_{\*} \alpha \cdot \beta$ for classes $\alpha \in CH^{\*}(X)$ and $\beta \in CH^{\*}(X)$.
Of course a similar results holds in cohomology if $f$ is a proper map of smooth manifolds, using Gysyn map for push-forward.
3) If $H < G$ are finite groups, we have two functors $\mathop{Ind}\_{H}^{G}$ and $\mathop{Res}\_{H}^{G}$, which can be seen as pull-back and push-forward maps between the representations rings $R(G)$ and $R(H)$. Again we have $\mathop{Ind}(U \otimes \mathop{Res} V) \cong \mathop{Ind} U \otimes V$.
Edit: one more example appears in the book linked in Peter's answer. It is a bit complicated to state, but basically (if I understand well)
4) for a compactly generated topological group $G$ and for $G$-spaces $A$ and $B$ one considers the category $G \mathcal{K}\_A$ of $G$-spaces over $A$ with equivariant maps (up to homotopy). Then for a $G$-map $f \colon A \to B$ one has functors $f^{\*} \colon G\mathcal{K}\_B \to G\mathcal{K}\_A$ and $f\_{!} \colon G\mathcal{K}\_A \to G\mathcal{K}\_B$ satisfying $f\_{!}(f^{\*}Y \wedge\_A X) \cong Y \wedge\_B f\_{!} X$.
There are probably several other variations which now I fail to recall. I should mention that in some situations 2) can be obtained by 1), but not always, as far as I know.
>
> Is there a unifying principle (even informal) which explains why in these diverse settings we should always have the same formula?
>
>
>
| https://mathoverflow.net/users/828 | Ubiquity of the push-pull formula | [This paper](http://www.math.uiuc.edu/K-theory/0573/) by Fausk, Hu and May does not exactly tell you why those maps should be isomorphisms in more concrete situations, but it cleanly explains the abstract settings in which they arise - look e.g. at Propositions 2.4 and 2.8 for equivalent formulations of projection formulas.
For an example of a projection formula that is not on the list in your question see equation 2.2.5 in [this book](http://www.math.uiuc.edu/K-theory/0716/) by May and Sigurdsson - it is an example for the abstract "Wirthmüller context" from the paper above, which, I think, inspired the authors to do the abstract analysis in the first place.
| 7 | https://mathoverflow.net/users/733 | 18814 | 12,535 |
https://mathoverflow.net/questions/18813 | 19 | This question may end up [closed], but I'm going to ask and let the people decide. It's certainly the kind of question that I'd ask people at tea, and it's not one whose answer I've been able find with Google.
TeX, I have heard, is Turing complete. In theory, this means that we can do modular arithmetic with LaTeX programs. I'd like to know how this can be done in practice.
Background: I've been using the \foreach command in TikZ to draw NxN arrays of nodes, indexed by pairs of integers (m,n). I'd like to be able to use modular arithmetic and an ifthenelse statement to put different decorations on the nodes, depending on the value of (m+n) mod p. Obviously, one *can* just do this by hand. But that's not the world I want to live in.
| https://mathoverflow.net/users/35508 | Modular Arithmetic in LaTeX | Get a current version of Ti*k*Z and use `\pgfmathmod`!
| 25 | https://mathoverflow.net/users/1409 | 18815 | 12,536 |
https://mathoverflow.net/questions/18817 | 9 | Is it true that for every integer $r$, the equation $2^m = 3^n + r$ has at most a finite number of integer solutions? I understand that this is a special case of Pillai's conjecture, which is unsolved.
If the statement is true, then can we verify the finiteness of the solution set using modular arithmetic? To be precise, is the following proposition true?
$$\forall r,\ \exists M,\ \exists N,\ \forall m,n \ge N,\ \ 2^m \not\equiv 3^n + r \pmod{M}$$
I have verified the proposition for $0 \le r \le 12$, and found the least possible modulus $M(r)$ for each $r$ in this interval. Note that $M(r) = 2$ if $r$ is even.
$M(1) = 8$, $M(3) = 3$, $M(5) = 1088$, $M(7) = 1632$, $M(9) = 3$, $M(11) = 8$.
| https://mathoverflow.net/users/4758 | Does 2^m = 3^n + r have finitely many solutions for every r? | Yes, it is true that this kind of equation ax+by=c, where a,b,c are non-zero and fixed and x,y are allowed to only have prime factors in a finite set, has only finitely many solutions. This is a special case of Siegel's theorem on integral points on curves.
Your second question may be unknown in the generality you pose. It is interesting that it holds. A remark: if there is a solution to $2^m = 3^n + r$, then $2^{m+k\phi(M)} \equiv 3^{n+k\phi(M)} + r (\mod M)$ for all $k,M$ if $(M,6)=1$, so if $M$ exists in this case, then $(M,6)>1$. If there is no solution to the equation $2^m = 3^n + r$, then the existence of $M$ (with $N=0$) is a special case of a conjecture of Skolem.
T. Skolem: Anwendung exponentieller Kongruenzen zum Beweis der
Unlösbarkeit gewisser diophantischer Gleichungen., Avh. Norske Vid. Akad.
Oslo, 12 (1937), 1–16.
Another comment. There are no solutions when $r=11$ but $M=8$ doesn't work since $2^2 \equiv 3^2 + 11 \mod 8$. $M(11)=205$. (Edit: $M(11)=8$ is OK. I misunderstood the definition, see comments)
| 15 | https://mathoverflow.net/users/2290 | 18819 | 12,539 |
https://mathoverflow.net/questions/18794 | 44 | I recently got interested in game theory but I don't know where should I start.
Can anyone recommend any references and textbooks?
And what are the prerequisites of game theory?
| https://mathoverflow.net/users/3124 | How to start game theory? | "[A course in game theory](http://theory.economics.utoronto.ca/books/)" by Martin J. Osborne and Ariel Rubinstein is probably the standard more mathematical starting point. A more concise, more modern, and slightly CS-leaning text is "[Essentials of Game Theory -- A Concise, Multidisciplinary Introduction](http://www.gtessentials.org/)" by Kevin Leyton-Brown and Yoav Shoham.
| 14 | https://mathoverflow.net/users/4762 | 18829 | 12,548 |
https://mathoverflow.net/questions/18764 | 7 | I am looking for a reference for the following well-known fact: For any subdiagram $\Delta\_0$ of of the Dynkin diagram $\Delta=D(G)$
of an adjoint simple group $G$ over an algebraically closed field $k$, there exists a reductive subgroup of maximal rank $G\_0\subset G$
with Dynkin diagram $\Delta\_0$.
To be more precise, I am looking for a reference for a proof of the following well-known lemma:
Lemma 1. Let $G$ be an adjoint, connected, simple algebraic group with Dynkin diagram $\Delta=D(G)$
over an algebraically closed field $k$ of any characteristic.
Let $\Delta\_0$ be a subdiagram of $\Delta$
(that is, a subset $\Pi\_0$ of the set $\Pi$ of vertices of $\Delta$,
together with all the edges of $\Delta$ connecting pairs of vertices of $\Pi\_0$).
Then there exists a connected reductive $k$-subgroup of maximal rank $G\_0$ of $G$
such that the corresponding adjoint semisimple group $G\_0^{ad}$
has Dynkin diagram $\Delta\_0$.
I know a simple proof of Lemma 1, but I would prefer to give a reference rather than a proof.
The proof goes as follows. Let $T$ be a maximal torus of $G$, and let $R=R(G,T)$ be the root system,
then our $\Pi$ is a basis of $R$. Let $S$ be the subgroup of $T$ orthogonal to $\Pi\_0$,
then it is a subtorus of $T$ (because $G$ is adjoint).
Set $G\_0=C\_G(S)$, the centralizer of $S$ in $G$.
Then $G\_0$ is a connected reductive subgroup of $G$.
It is easy to see that (the adjoint group of) $G\_0$ has Dynkin diagram $\Delta\_0$.
Note that Lemma 1 is a special case of the following Lemma 2,
for which I would also be happy to have a reference.
Lemma 2. Let $G$ be an adjoint, connected, simple algebraic group
over an algebraically closed field $k$ of any characteristic.
Let $T$ be a maximal torus of $G$, and let $R=R(G,T)$ be the root system.
Let $R\_0$ be a closed symmetric subset of $R$.
Then there exists a connected reductive $k$-subgroup of maximal rank $G\_0$ of $G$
with root system $R\_0$.
I will be grateful to any references, comments, etc. (also to a proof of Lemma 2).
Mikhail Borovoi
| https://mathoverflow.net/users/4149 | Reductive subgroup corresponding to a subdiagram of the Dynkin diagram of a simple group | Dear Mikhail,
If I understand correctly, your Lemma 2 is implied by
SGA 3 Exposé 22, Théorème 5.4.7.
Everything is on a general base S (that you may take as your algebraically closed field). The kind of subgroup you want is called "de type (R)" (see Définition 5.2.1) and a subset of R that corresponds to such a group is also called "de type (R)". Now the theorem above exactly says that when a subset of R is closed, it is "de type (R)" which exactly means that there is a corresponding connected subgroup of G. By the way, Théorème 5.4.7 does not assume the subset to be symmetric, and you get things like Borel subgroups if you take only "half" of the roots. In the symmetric case, the group is reductive by Proposition 5.10.1.
Hope this helps.
Baptiste Calmès
| 10 | https://mathoverflow.net/users/4763 | 18830 | 12,549 |
https://mathoverflow.net/questions/18594 | 3 | Consider the following
$f\_{t+1}(z)=p\_{12} f\_{t}(z/A)+ p\_{21} f\_{t}(z/B)+p\_{22} f\_{t}(z/(A+B))$, where $A$, $B$, and the $p$'s are constants and $f\_t$ is a probability distribution. Are there any nice distribution families that are preserved under the transformation? Fail that, are there $f\_t$ such that $f\_{t+1}$ has a closed form?
It's motivated by the following problem:
Let there be two simple bonds that either default or pay off a return on investment (That may or may not be correlated), denote the bonds as random variables $Z\_1$ and $Z\_2$.
Now throw in a population of investors, with wealth following a distribution $W$, investing some fixed percentage of their income in the two bonds(investing a fixed percentage is a Nash equilibrium under the model I'm working with). The resulting after-investment wealth distribution will be a mixture of dilations of the original distribution, I'm trying to find a distribution to work with that will make things simple when studying the behavior of the system over time.
Any ideas?
| https://mathoverflow.net/users/4126 | Finding a distribution family that is preserved under mixture. | One can rewrite the problem in terms of products of i.i.d. random variables as follows.
Assume that $X\_t$ has distribution density $f\_t$. Then, the relation between $f\_t$ and $f\_{t+1}$ means that one can choose $X\_{t+1}=X\_tZ\_{t+1}$, where the $Z\_t$ are i.i.d. and $Z\_t=A$ or $B$ or $A+B$, with probabilities $p\_{12}A$, $p\_{21}B$ and $p\_{22}(A+B)$, respectively.
Hence, for the relation between $f\_t$ and $f\_{t+1}$ to make sense, one must assume that the three nonnegative numbers $p\_{12}A$, $p\_{21}B$ and $p\_{22}(A+B)$ sum to $1$, and when this is so, $X\_t=X\_0Z\_1Z\_2\cdots Z\_t$.
This tells you that:
* $E(X\_t)=E(X\_0)m^t$ for every $t$, with $m=E(Z\_1)$, that is,
$$
m=p\_{12}A^2+p\_{21}B^2+p\_{22}(A+B)^2.
$$
* $t^{-1}\log X\_t$ converges almost surely to $\mu=E(\log Z\_1)$, that is,
$$
\mu=p\_{12}A\log A+p\_{21}B\log B+p\_{22}(A+B)\log(A+B).
$$
* $\log X\_t$ follows the multinomial distribution with parameters $t$ and $(p\_{12}A,p\_{21}B,p\_{22}(A+B))$, or, more precisely, the convolution of this multinomial with the distribution of $\log X\_0$.
Unfortunately, these remarks do not help much if one is interested in closed form formulas. Sorry.
| 2 | https://mathoverflow.net/users/4661 | 18835 | 12,552 |
https://mathoverflow.net/questions/18844 | 6 | To elaborate a bit, I should say that the question of the existence of a complete metric is only of interest in the case of manifolds of infinite topological type; if a manifold is compact, any metric is complete, and if a noncompact manifold has finite topological type(ie is diffeomorphic to the interior of a compact manifold with boundary,) one can contruct a complete metric on the manifold with boundary via a partition of unity, and then divide by the square of a defining function to get an complete asymptotically metric on the interior.
I have absolutely no intuition for how "wild" these manifolds can be. The only examples I can think of are infinite connected sums and quotients of negatively curved symmetric spaces by sufficiently complicated groups, but I'd imagine that one can construct some pathological examples by limiting arguments.
| https://mathoverflow.net/users/2497 | Does every smooth manifold of infinite topological type admit a complete Riemannian metric? | By Whitney embedding theorem any smooth manifold embeds into some Euclidean space as a closed subset. The induced metric is complete.
In fact, a good exercise is to show that any Riemannian metric is conformal to a complete metric.
| 18 | https://mathoverflow.net/users/1573 | 18851 | 12,561 |
https://mathoverflow.net/questions/18798 | 2 | This question is related to [this previous question](https://mathoverflow.net/questions/18758/drawing-a-combinatorial-3-configuration-of-points-and-lines-with-pseudolines). Many combinatorial configurations have Levi graphs which may be represented as derived graphs obtained from voltage graphs over a cyclic group; in a number of such cases, it is possible to represent the combinatorial configuration as a geometric configuration (i.e., using points and straight lines in the Euclidean plane).
Given a bipartite graph which is obtained from a voltage graph, we can view it as a Levi graph of some combinatorial configuration. Is it possible to draw all such configurations using pseudolines? If not, are there easy/known constraints on the ones that fail? (e.g., if there are more than x points in the configuration, then things work? You can't use such-and-so groups as the cyclic group for the voltage graph?)
(Does the Heawood graph have a voltage-graph representation? If so, it makes the first question easy to answer, but the second one is still interesting. Maybe.)
| https://mathoverflow.net/users/607 | Are combinatorial configurations whose Levi graphs may be represented as covering graphs over voltage graphs realizable with pseudolines? | The fist part of your question has a negative answer, since both Fano plane (73) and Moebius-Kantor configuration (83) are *cyclic configurations*. [Here](http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6V00-4520GM0-F&_user=10&_coverDate=02%252F06%252F2002&_alid=1258918945&_rdoc=32&_fmt=high&_orig=search&_cdi=5632&_sort=r&_docanchor=&view=c&_ct=39&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=966cc8a16ea0603a5f1aa4ac9044ccd2) it is shown that the cyclic covering graphs over a dipole with girth at least 6 are exactly Levi graphs of combinatorial cyclic configurations. In your terminology, each Levi graph of a cyclic configuration has a "voltage-graph representation". Note: a *dipole* is a graph consisting of two vertices and a number of parallel edges between them. In particular both the Heawood graph and the Moebius-Kantor graph are counterexaples.
If v = uw is a composite number, it is sometimes possible to find a suitable voltage graph on 2u vertices and voltages from the cyclic group Zw to produce a straight-line drawing of the corresponding [polycyclic configuration](http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6WDY-48BKR3H-6&_user=10&_coverDate=05%252F31%252F2003&_alid=1258918945&_rdoc=35&_fmt=high&_orig=search&_cdi=6779&_sort=r&_docanchor=&view=c&_ct=39&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=144ff4ac2004d0e709e163b00381794d) (v3). Unfortunately this method does not apply to cyclic configurations (vk), for v prime and is not certainly not understood well for k > 4. I am not sure what happens if pseudolines are admitted in such drawings having *rotational symmetry*.
| 1 | https://mathoverflow.net/users/4400 | 18858 | 12,567 |
https://mathoverflow.net/questions/18846 | 1 | I can't understand this sentence i the article of Majid "Tannaka-Krein theorem for quasi-Hopf algebras and other results" about the reconstruction of a quasi-algebra (in fact its dual) from a given braided category (C,c) with a forgetful functor F ; at the end it considers the action of Drinfeld-twisting upon quasi-algebra and says:
if A (quasi-algebra) corresponds to (C,F,c) then A\_twisted corresponds to (C,F,c') for a new choice of braiding c'.
i am very disturbed: the associativity constraint is modified too obviously; sorry for my very naive question.
| https://mathoverflow.net/users/4770 | Why does twisting quasi-Hopf algebras work (as in majid's article) | You might find these notes helpful as a second source to read fro:
<http://www-math.mit.edu/~etingof/tenscat1.pdf>
There is a careful elaboration of the various variants of Tannaka-Krein construction, and the case of quasi-bialgebras and quasi-Hopf algebras is discussed. So far as I recall, Drinfeld twisting of a quasi bi-algebra just means acting on the original category by a tensor functor which is the identity on objects (and so is only interesting as a tensor functor).
As such, it does have a fairly straightforward effect on the associator $\phi$. If you twist by an invertible element $F\in H\otimes H$, then the associator $\phi^F$ is something like $\phi^F:=B\phi B^{-1}$, where $B:=F(\Delta\otimes\operatorname{id})(F)$, or maybe $F(\operatorname{id}\otimes\Delta)(F)$, I can't remember.
It might help you to note that nothing in your question relies on the braiding; there is a reconstruction theorem for any (quasi)bi-algebra with a fiber functor, and since you're asking about the associator, it isn't too important that there's a braiding lying around.
What leads you to expect some more complicated effect on the associator?
| 3 | https://mathoverflow.net/users/1040 | 18861 | 12,568 |
https://mathoverflow.net/questions/18854 | 3 | The classification theorem for surfaces says that the complete set of homeomorphism classes of surfaces is
{ $S\_g : g \geq 0$ } $ \cup$ { $N\_k : k \geq 1$ },
where $S\_g$ is a sphere with $g$ handles, and $N\_k$ is a sphere with $k$ crosscaps. The first homology groups are easy to compute. They are $H\_1 (S\_g) = \mathbb{Z}^{2g}$, and $H\_1 (N\_k)=\mathbb{Z}^{k-1} \times \mathbb{Z} / 2\mathbb{Z}$. My question concerns how the homology groups change once we start cutting holes in our surface.
In the orientable case, it is easy to see what happens. The first hole that we cut out does not change the homology. Every additional hole then introduces another factor of $\mathbb{Z}$.
In the non-orientable case something peculiar happens. Consider the projective plane with homology group $\mathbb{Z} / 2 \mathbb{Z}$. If I cut out a hole, then I get the Mobius strip, which has homology group $\mathbb{Z}$ (it is homotopic to a circle). In general, if I cut a hole out of $N\_k$, then in the homology group I lose a factor of $\mathbb{Z} / 2 \mathbb{Z}$, and introduce a factor of $\mathbb{Z}$. Each additional hole will then just introduce another factor of $\mathbb{Z}$.
**My question:** In the non-orientable case what happened to the factor of $\mathbb{Z} / 2 \mathbb{Z}$? Is there a nice geometric explanation of why it went away? I'm slightly disturbed because I had the intuition that torsion was supposed to record non-orientability, but I guess this doesn't work for surfaces with holes.
| https://mathoverflow.net/users/2233 | Homology of Surfaces with Holes | Autumn's answer captures the essence why there is a $\mathbb{Z}\_2$ is in the first homology
of a closed nonorientable surface.
If you remove a disk from a closed surface, the resulting object has a $1$-dimensional $CW$-complex as a strong deformation retract, so that the homology of the resulting object
has no torsion.
A closed nonorientable surface is always the result of the connect sum of an orientable surface
with a projective plane or two projective planes. That is you are gluing one Moebius band, or two Moebius bands into the boundary of an orientable surface. The core or cores of the Moebius bands, oriented appropriately represent the generator of the $2$-torsion in the first homology of the surface.
| 6 | https://mathoverflow.net/users/4304 | 18862 | 12,569 |
https://mathoverflow.net/questions/10730 | 8 | There is a textbook theorem that the categories of unipotent algebraic groups and nilpotent finite-dimensional Lie algebras are equivalent in characteristic zero. Indeed, the exponential map is an algebraic isomorphism in this case and the group structure can be defined in terms of the Lie algebra structure and vice versa via the Campbell-Hausdorff series, which is finite due to nilpotency.
My problem is that I am unable to locate any textbook where this textbook theorem is stated. The books by [Borel](http://books.google.com/books?id=R31Z75Yvaj8C&printsec=frontcover&dq=Linear+algebraic+groups&cd=1#v=onepage&q=&f=false), [Humphreys](http://books.google.com/books?id=hNgRLxlwL8oC&printsec=frontcover&dq=Linear+algebraic+groups&cd=2#v=onepage&q=&f=false), [Springer](http://books.google.com/books?id=IVGvJ17FnMsC&printsec=frontcover&dq=Linear+algebraic+groups&cd=3#v=onepage&q=&f=false), [Serre](http://books.google.com/books?id=hasbV48Pm_AC&pg=PP1&dq=serre+lie+algebras+and+lie+groups&cd=1#v=onepage&q=&f=false) do not seem to mention this theorem.
The only reference I was able to locate is this [original paper](http://www.jstor.org/stable/2037665) by Hochschild (which refers to his earlier papers), but he does it in a heavy Hopf-algebra language that is good, too, but still leaves one desiring to find also a simple textbook-style exposition. Later Hochschild wrote a book "Basic Theory of Algebraic Groups and Lie Algebras" on the subject, to which I have presently no access, but judging by Parshall's [review](http://projecteuclid.org/euclid.bams/1183551302), it is certainly not textbook-style.
Could anyone suggest a simple reference for this textbook theorem?
| https://mathoverflow.net/users/2106 | References for theorem about unipotent algebraic groups in char=0? | Demazure-Gabriel, *Groupes algebriques*, Tome I (published in 1970) is a more explicit source, if available. Chapitre IV treats "groupes affines,
nilpotents, resolubles", while Chapitre V specializes to commutative affine groups.
Typically they work over an (almost) arbitrary field $k$, but IV.2.4 is
devoted to "groupes unipotents en caracteristique 0". This seems to be as
full an account as you will find in a textbook; see especially Corollaire
4.5 for the category equivalence you want.
Later textbooks in English
including Hochschild focus mainly on the structure/classification of reductive
rather than arbitrary affine groups. Even this much of the story told without scheme language is fairly long for graduate courses. It's regrettable from the reference viewpoint that Demazure-Gabriel gave up after one volume.
| 7 | https://mathoverflow.net/users/4231 | 18869 | 12,575 |
https://mathoverflow.net/questions/18840 | 23 | I wish to learn nonstandard analysis. Are there any good book recommendations? I'm familiar with the ZFC system, and learnt analysis the classical way. I've found some undergraduate texts, but they are too verbose.
If there are applications to complex or functional analysis, that would be great.
Thanks in advance.
| https://mathoverflow.net/users/nan | nonstandard analysis book recommendation | This one sounds like what you want:
[Arkeryd, Cutland and Henson: Nonstandard Analysis, Theory and Applications.](http://books.google.co.uk/books?id=X0BbnEP3jN0C&pg=PA1&lpg=PA1&dq=nonstandard+analysis+cutland&source=bl&ots=B_xdfk9wsQ&sig=iIZA61f8iIvbpE1Wdap0azWtC-c&hl=en&ei=rx6lS_3hEo2OjAeX8tnzCQ&sa=X&oi=book_result&ct=result&resnum=1&ved=0CAsQ6AEwAA#v=onepage&q=&f=false)
I took a course as undergraduate which followed (parts of) this book - I first accompanied it with the more friendly written Goldblatt to get some feeling for the subject, then switched to this one, when I also started finding Goldblatt too "verbose". It found it very well readable.
Cutland has produced other enjoyable writings and there also is [Loeb, Wolff: Nonstandard analysis for the working mathematician](http://books.google.co.uk/books?id=jcjBpRnMiecC&pg=PA168&lpg=PA168&dq=nonstandard+analysis+cutland&source=bl&ots=b4huM_RoNF&sig=PSX8vX6Jl7WWtJqDuptVAfid1qM&hl=en&ei=ih-lS5z2HIvNjAeH1JnuCQ&sa=X&oi=book_result&ct=result&resnum=10&ved=0CCUQ6AEwCTgK#v=onepage&q=nonstandard%2520analysis%2520cutland&f=false), which I haven't read but which follows a very similar agenda at a very similar pace, according to the table of contents.
Enjoy!
| 10 | https://mathoverflow.net/users/733 | 18872 | 12,578 |
https://mathoverflow.net/questions/9915 | 14 | If G is a reductive algebraic group (say over ℂ), T a maximal torus, then we can consider its Weyl group W which acts on the abelian group Y of one parameter subgroups of T. Thus we may form the semidirect product, which I will call the affine Weyl group.
In the semisimple simply connected case, this affine Weyl group is a Coxeter group. In the general setup I have here, the affine Weyl group is no longer a Coxeter group, but shares some properties of Coxeter groups. For example one can still define a length function, and considering an Iwahori subgroup of a loop group, we get something that looks like it wants to be a BN-pair/Tits system (but isn't, since we don't have a Coxeter group).
Since the axiomatic setup of Coxeter groups and BN-pairs is so convenient, I ask if there has been developed a generalisation of this setup to include the affine Weyl groups I mention above. I have some vague feeling that this should include some sort of 'disconnected' (I am thinking in the algebraic group sense when I use this word) Coxeter group, where the connected component is a genuine Coxeter group.
| https://mathoverflow.net/users/425 | Affine Weyl groups as Coxeter groups | In the abstract Bourbaki set-up, the affine Weyl group is defined to be a
semidirect product of an *irreducible* Weyl group with its coroot lattice.
This is naturally a Coxeter group, characterized in terms of its
positive semidefinite Coxeter matrix. The basic theory is developed
independently of applications in Lie theory, but is directly usable if you
start with a connected semisimple algebraic group (over an algebraically
closed field) and require its root system to be *irreducible* of type A, B,
etc. Most of the time this causes no trouble. While it is natural to work
with a connected reductive group, people often use the expression "affine
Weyl group" too loosely in this general context. For example, the standard
features of alcove geometry require irreducibility. Otherwise you have
to deal with products of simplexes, etc. In any case, the difference between
reductive and semisimple groups such as general linear and special linear
is sometimes significant.
In the Iwahori-Matsumoto (or Bruhat-Tits) setting over local fields, a
more intrinsic affine Weyl group occurs directly within the structure of
the group itself. Here one has to be cautious about applying
abstract Coxeter group theory or BN-pair theory, as I believe most authors
are. Already in the proceedings of the 1965 Boulder AMS summer institute,
Iwahori had to formulate a more complicated "generalized BN-pair" formalism
for this situation. I'm not sure what has become standard by now in the
literature.
In other situations (the classical study of compact Lie groups, or
the later application of affine Weyl groups in modular representation
theory starting with Verma) there is usually no difficulty about specializing
to the irreducible case. Here the affine Weyl group lives outside the
actual group under study. This is the situation I'm most comfortable with.
You need to make precise the setting in which you really want to study reductive groups, in order to adapt the Bourbaki language and results.
There are several distinct issues here: 1) Extra care is needed in treating
disconnected algebraic groups such as orthogonal groups, or in treating
reductive rather than semisimple groups. 2) Adjoint groups, simply
connected groups, and the occasional intermediate type: not all details of
structure are exactly the same. 3) Most important for working over
local fields is the natural use of an "extended affine Weyl group" (as in
much of Lusztig's work involving Hecke algebras, cells, etc.). Here you
start with the Bourbaki version of the affine Weyl group (a Coxeter group)
and form a semidirect product with a finite group $\Omega$ isomorphic to the weight lattice mod root lattice (universal center). This amounts to
working with a semidirect product of the Weyl group and the full (co)weight
lattice rather than the (co)root lattice. Fortunately it's easy to extend
notions like length function to this extended group.
EDIT: Besides Bourbaki's treatment of Coxeter groups and root systems (1968),
foundational papers from that era include Iwahori-Matsumoto (IHES Publ. Math. 25, 1965), at <http://www.numdam.org>, and Iwahori's 1965 article in the AMS Boulder proceedings
<http://www.ams.org/books/pspum/009/0215858/pspum0215858.pdf>, followed by much more technical work by Bruhat-Tits.
Lusztig has written many technical papers on *extended* affine Weyl groups and corresponding affine Hecke algebras, including his four part series on cells in affine Weyl groups and later work on multiparameter cases. Much of this work is motivated by reductive groups over local fields, as well as the modular representation theory of reductive groups and their Lie algebras (where "linkage" of weights appears at first relative to an extended affine Weyl group).
| 14 | https://mathoverflow.net/users/4231 | 18875 | 12,580 |
https://mathoverflow.net/questions/18876 | 2 | There is the invariant Maurer–Cartan 1-form on a compact semi-simple Lie group G. So if we pull it back using a map from X to G then we get a G-connection on X. The question is, can all G-connections on X ( here just regarded as elements in ${\Omega}^1(X, \mathfrak{g})$, i.e. we are using the trivial G-bundle over X ) arise in this way?
For flat connections, I believe the answer is yes, which can be seen when one "cut" the Riemann surface into a disc with boundary conditions. But I have no idea about the general case.
Some facts that may be useful ( when I tried to figure out an answer ) :
1. It is well-known that all G-connections on X can arise from a map from X to BG by pulling back a special connection on BG.
2. If ${\pi}\_1(G) = {\pi}\_2(G) = 0$, then I guess there shall be no homotopic non-trivial map from X to G ( note ${\pi}\_n(X) = 0$ when $n > 2$ ), and hence all maps from X to G are "integrable" in the sense that they can be induced by applying the exponential map on maps from X to $\mathfrak{g}$. Not sure about how this will help though.
3. Actually I am not even sure about the $G = U(1)$ case. On the other hand, if the answer is true, then probably the curvature of the induced connection will have some topological constraints coming from the map ${\pi}\_1(X) \rightarrow \mathbb{\pi}\_1(G)$.
I haven't touch math for quite some time due to personal reasons, so I hope I am not making any trivial mistakes in this post ;-)
| https://mathoverflow.net/users/4782 | Can all G-connections on a Riemann surface X be induced by maps from X to G | José Figueroa-O'Farrill has already pointed out one necessary condition, namely that your connection must be flat. The remaining condition is that the monodromy should be trivial. In what follows $X$ is any connected smooth manifold, not necessarily a surface, and $G$ is any Lie group.
Let's first consider the analogous situation when $G$ is replaced by $\mathbb{R}$. You can think of a one-form $\omega\in \Omega^1(X;\mathfrak{g})$ as potentially being the derivative of a map $X\to G$, just as a one-form $\eta\in \Omega^1(X;\mathbb{R})$ is potentially the derivative of a map $X\to \mathbb{R}$. We want to know when these really are the derivative of some map, i.e. when we can integrate these forms. (You mentioned the exponential map, but I think integration is the right metaphor here.)
There is a local obstruction, namely that if $\eta$ is to be integrable (meaning $\eta=df$ for some $f$) it must be closed, meaning $d\eta=0$; the Poincaré lemma tells us this is a sufficient condition for $\eta$ to be locally integrable. Then there is also a global condition, that the integral of $\eta$ around every closed loop must be 0 (unlike $d\theta$ on the circle, which has integral $2\pi$); Stokes' theorem tells us this is a necessary condition for $\eta$ to be globally integrable. If we have these conditions, recovering the map $f$ from $\eta$ is easy; just write $f(p)=\int\_\ast^p \eta$, which is well-defined by the above two conditions.
Now let's try to do the same for $\mathfrak{g}$-valued one-forms. Start with a connection on the trivial $G$-bundle $X\times G$ with connection form $\omega\in \Omega^1(X;\mathfrak{g})$. We've talked about the connection being flat, which means that $d\omega+\frac{1}{2}[\omega,\omega]=0$; but what does that have to do with flatness or integrability? Well, you can show that $d\omega+\frac{1}{2}[\omega,\omega]$ measures the Lie bracket of two horizontal vector fields, or rather measures the vertical part of the Lie bracket. Thus if this vanishes, the bracket of two horizontal vector fields is horizontal. By the Frobenius integrability theorem, this implies that the horizontal distribution of the connection is integrable; another way to say this is that parallel transport is locally well-defined. Now pick a basepoint $\ast$ and restrict your attention to a small neighborhood $U$ of $\ast$. Since parallel transport is well-defined on $U$, we get a function $T\colon U\to G$ by saying that the parallel transport from $\ast$ to $u$ (along any path) is multiplication by $T(u)$.
**Key point:** if you pull back the tautological form on $G$ by this "parallel transport" map $T$, the form you get is the same as your original $\omega$!
What this tells us is that if a flat connection on $X\times G$ comes from a map $f:X\to G$, then you can recover $f$ by looking at the parallel transport of the connection. (The analogue is that if $\eta=f^\ast(dx)$ for some $f\colon X\to \mathbb{R}$, you can recover $f$ by integrating $\eta$, also known as the fundamental theorem of calculus.) Thus flatness, in the form of the Maurer-Cartan equation, is the local obstruction to integrability; here the Frobenius integrability theorem plays the role that the Poincaré lemma does in the real case. To prove the key point is really just a matter of definitions: think about the correspondence between a connection, its connection form, and its parallel transport.
In particular, this tells us that parallel transport must be not just locally well-defined, but globally well-defined (meaning independent of the path), since transport along any path from $\ast$ to $p$ is always multiplication by $f(p)\in G$. The monodromy of a flat connection is the map $\pi\_1(X,\ast)\to G$ which sends a loop to the parallel transport around that loop, and so another way to say "parallel transport is globally well-defined" is that the monodromy is trivial.
This can all be summed up by saying that if $X$ is simply connected, we have an on-the-nose bijection $C^\infty(X,G)\longleftrightarrow \{\omega\in \Omega^1(X;\mathfrak{g})\vert d\omega+\frac{1}{2}[\omega,\omega]=0\}$. (Here on the left we assume the maps take the basepoint $\ast\in X$ to $1\in G$.) If $X$ has fundamental group, we need to add on the right side the additional condition that the monodromy of $\omega$ be 0. This is hard to write down just in terms of $\omega$, but for the corresponding connection it is just that parallel transport is totally path-independent.
| 7 | https://mathoverflow.net/users/250 | 18885 | 12,586 |
https://mathoverflow.net/questions/18893 | 6 | The idea hit me when I was in my Elliptic Curve Cryptography class. $Z\_n \leftrightarrow Z\_{f\_1} \times Z\_{f\_2} \times ...$ where $f\_1 \times f\_2 \times ... = n$ and $\{f\_1, f\_2, ...\}$ are pairwise coprime. Applications of this Chinese Remainder Theorem not only include computational speedups (in the case of decryption in RSA) but also stronger cryptographic attacks against $Z\_n$ (for example, Pollard Rho factoring exploits this structure). Can we extend these applications into other areas? (Admittedly I don't know many computationaly examples where this could be useful, but can imagine that Mathematica/Maple would find some uses).
So the real question: is this property just a "coincidence" or are there analogs in other groups? If there are, is there some group theory analog that applies equally well to every group? If there are not, what underlying structure in the natural numbers makes this possible?
| https://mathoverflow.net/users/3737 | Analog to the Chinese Remainder Theorem in groups other than Z_n. | The Chinese Remainder theorem is usually thought of as an isomorphism of *rings*, not just of cyclic groups. In this regard it has a vast generalization:
Theorem (Ideal-theoretic CRT): Let R be a commutative ring, and let $I\_1,\ldots,I\_n$ be a finite set of ideals in $R$ which are **pairwise comaximal**: for all $i \neq j$, $I\_i + I\_j = R$. Then $I\_1 \cap \ldots \cap I\_n = I\_1 \cdots I\_n$ and the natural homomorphism
$R/I\_1 \cdots I\_n = R/I\_1 \cap \ldots \cap I\_n \rightarrow \bigoplus\_{i=1}^n R/I\_i$
is an isomorphism. (See e.g. Theorem 41 on p.31 of [http://alpha.math.uga.edu/~pete/integral.pdf.](http://alpha.math.uga.edu/%7Epete/integral.pdf.))
One could also think of $\mathbb{Z}/n\mathbb{Z}$ as a $\mathbb{Z}$-module, and then the CRT decomposition is a special case of primary decomposition for $R$-modules. In general rings, primary decomposition is somewhat complicated (e.g. it need not be unique), but for finitely generated torsion modules over a PID there is a straightforward analogue.
Finally, thinking about it in terms of groups, CRT has the following generalization: a finite group is **nilpotent** iff each Sylow $p$-subgroup is normal and $G$ is the direct product of its Sylow $p$-subgroups. There are Sylow decompositions in certain other group-theoretic contexts as well, e.g. nilpotent profinite groups.
| 20 | https://mathoverflow.net/users/1149 | 18899 | 12,596 |
https://mathoverflow.net/questions/18877 | 18 | 1. Does ${\mathbb R}$ have proper, countable index subrings? By countable I mean finite or countably infinite. By subring I mean any additive subgroup which is closed under multiplication (I don't care if it contains $1$.) By index, I mean index as an additive subgroup.
2. Given some real number $x$, when is it possible to find a countable index subring of ${\mathbb R}$ which does not contain $x$?
| https://mathoverflow.net/users/4783 | Are there countable index subrings of the reals? | Perhaps surprisingly, it turns out that such subrings do exist. This was proved in Section 2 of my paper:
Simon Thomas, Infinite products of finite simple groups II,
J. Group Theory 2 (1999), 401--434.
The basic idea of the proof is quite simple. Clearly the ring of $p$-adic integers has countable index in the field of $p$-adic numbers. Now the $p$-adic integers are the valuation ring of the obvious valuation on the field of $p$-adic numbers ... and it turns out to be enough to show that $\mathbb{C}$ has an analogous valuation. This is true because $\mathbb{C}$ is isomorphic to the field of Puiseux series over the algebraic closure of $\mathbb{Q}$, which has an appropriate valuation.
| 28 | https://mathoverflow.net/users/4706 | 18900 | 12,597 |
https://mathoverflow.net/questions/18903 | 7 | In the article ['André Weil As I Knew Him'](http://www.ams.org/notices/199904/shimura.pdf) in the April 1999 issue of Notices of the AMS, Shimura recounts how in 1996 André Weil (then 90 years old) didn't remember a mistake of Minkowski. Specifically, quoting from mid-paragraph:
*"... In fact, to check that point, I asked him whether Minkowski was reliable. He said, 'I think so.' At that point I realized that his recollection was faulty, since Minkowski gave an incorrect formula, as Siegel pointed out, and that was known to most experts. ..."*
My question is:
**Where specifically in Siegel and Minkowski's published works can I find the aforementioned formula (of Minkowski) and observation (of Siegel)?**
| https://mathoverflow.net/users/2604 | What is Shimura referring to by "an incorrect formula given by Minkowski... known to most experts." | I believe he is referring to the result in Minkowski's dissertation that gives a formula for the mass (in the number/lattice theory sense) over a genus, that is, the sum of reciprocals of the group orders of all inequivalent quadratic forms in a genus. ([wikipedia](http://en.wikipedia.org/wiki/Smith%E2%80%93Minkowski%E2%80%93Siegel_mass_formula))
Siegel found and corrected the error in Minkowski's formula and also made many more generalizations in [this Annals paper](http://www.jstor.org/pss/1968644). Namely, there is an incorrect power of 2 in the formula.
| 12 | https://mathoverflow.net/users/239 | 18907 | 12,602 |
https://mathoverflow.net/questions/18916 | 6 | Global fields consist of finite extensions of $\mathbb{Q}$ (algebraic number fields) and finite extensions of $\mathbb{F}\_q(x)$ (function fields in 1 variable over a finite field). The latter are isomorphic to the category of curves over $\mathbb{F}\_q(x)$, and they can be generalized to function fields in $n$ variables over $\mathbb{F}\_q(x)$, which are isomorphic to the category of varieties over $\mathbb{F}\_q(x)$. Is there an analogous generalization of algebraic number fields?
| https://mathoverflow.net/users/4692 | Can algebraic number fields be generalized in a similar way to function fields in 1 variable over a finite field? | The function fields (in one or more variables) over $\mathbb{F}\_q$ are precisely the infinite, finitely generated fields of characteristic $p$. Thus an at least reasonable characteristic $0$ analogue is given by the (necessarily infinite!) finitely generated fields of characteristic $0$. In other words, function fields in finitely many (possibly zero) variables over a number field $K$.
Indeed there has been much work on generalizing arithmetic geometric statements over global fields to arithmetic geometric statements over arbitrary infinite, finitely generated fields. The one which springs most readily to my mind is the following generalization of the Mordell-Weil theorem due to Lang and Neron: the group of rational points on any abelian variety over any finitely generated field is a finitely generated abelian group.
| 5 | https://mathoverflow.net/users/1149 | 18917 | 12,609 |
https://mathoverflow.net/questions/18926 | 6 | Let $X\_\bullet \longrightarrow Y\_\bullet \longleftarrow Z\_\bullet$ be a diagram of simplicial spaces (=bisimplicial sets, if you like).
On p. 14-9 of [these notes](http://www.math.washington.edu/~warner/TTHT_Warner.pdf) there is an example which shows that if $Y\_\bullet$ and $Z\_\bullet$ are levelwise connected then the homotopy pullback of the geometric realisation of the diagram is the geometric realisation of the levelwise homotopy pullback.
The theorem is proved using the Bousfield-Friedlander theorem, which only requires that $\pi\_0Z\_\bullet \to \pi\_0Y\_\bullet$ is a Kan fibration and that $Z\_\bullet$ and $Y\_\bullet$ are $\pi\_\*$-Kan. Being levelwise connected implies both of these conditions, but is not necessary.
Can the conditions of the Bousfield-Friedlander theorem be relaxed? How about if $Z\_\bullet \to Y\_\bullet$ is something like a $\pi\_\*$-Kan fibration", though I'm not sure of the precise definition this should have?
| https://mathoverflow.net/users/318 | Homotopy pullbacks of simplicial spaces, and Bousfield-Friedlander | The following represents what I know about this; I don't know of a published reference.
Given a map $f:Z\_\bullet\to Y\_\bullet$ of simplicial "spaces" (to make things easy, assume spaces are simplicial sets), let's call it a realization quasi-fibration (RQF) if for every $U\_\bullet\to Y\_\bullet$, the homotopy pullback of the geometric realizations is weakly equivalent to the realizations of the levelwise homotopy pullbacks. The Bousfield-Friendlander theorem gives a sufficient condition for $f$ to be a RQF, in terms of the dreaded $\pi\_\*$-Kan condition.
Some facts:
* The pullback of an RQF $f$ along any $U\_\bullet\to Y\_\bullet$ is itself an RQF.
* Let $F[n]\_{\bullet}$ be the simplicial space which is free on a point in degree $n$. Then $f$ is an RQF if and only if its pullback along all $g: F[n]\_\bullet \to Y\_\bullet$, for all $n$, is an RQF.
These two facts are consequences of something that is sometimes called "descent"; basically, the facts that homotopy colimits distribute over homotopy pullbacks, and compatible homotopy pullbacks assembled by a homotopy colimit result in a homotopy pullback.
So the above gives exact criteria for $f$ to be an RQF. Whether the pullback of an RQF $f$ along a map $g$ is again an RQF only depends on the homotopy class of $g$. So if $f:Z\_\bullet\to Y\_\bullet$ is any map, let
$\pi\_0Y$ be the simplicial set whose $k$-simplices are $\pi\_0(Y\_k)$, which is to say all homotopy classes of maps $F[k]\_\bullet\to Y\_\bullet$. Let $RQF(f)\subseteq \pi\_0Y$ be the sub-simplicial set whose $k$-simplices correspond to $g:F[k]\_\bullet\to Y\_\bullet$ such that the pullback of $f$ along $g$ is an RQF.
So the criterion is: $f$ is an RQF iff $RQF(f)=\pi\_0Y$.
It turns out that since geometric realization always preserves products, any map $Z\_\bullet \to point\_\bullet$ is an RQF. Thus $RQF(f)$ contains all $0$-simplices of $\pi\_0Y$. Thus, if all $Y\_k$ are connected, $f$ is an RQF, which implies the result you describe.
| 7 | https://mathoverflow.net/users/437 | 18927 | 12,614 |
https://mathoverflow.net/questions/18935 | 1 | I'm trying to teach myself category theory from Steve Awodey's [Category Theory](http://www.andrew.cmu.edu/course/80-413-713/notes/). Chapter 2 asserts:
>
> It is not hard to see that a filter F is an ultrafilter just if for every element b ∈ B, either b ∈ F or ¬b ∈ F, and not both (exercise!).
>
>
>
I've managed to prove the backwards implication, but the forwards implication is eluding me. I'm guessing the correct approach is to consider a filter F such that there exists a b ∈ B such that neither b ∈ F or ¬b ∈ F, and construct a superset filter F' which contains b, but I can't figure out how to construct F' and prove that it's a filter. Any hints much appreciated!
| https://mathoverflow.net/users/4793 | F is ultrafilter over a Boolean algebra implies that for every b, either b or not-b is in F? | Assume you have a proper filter $F$ that avoids both $b$ and $\neg b$. Then, you could consider the filter generated by $F\cup\{b\}$ - which is to say the smallest filter $F'$ containing $F$ and $b$.
Since $F$ was a proper filter it follows that $0\not\in F$.
If $0\in F'$, then this means that there is some $f\in F'$ such that $b\wedge f = 0$. Now, $\neg b=0\vee\neg b=(b\wedge f)\vee\neg b=(b\vee\neg b)\wedge(f\vee\neg b)=1\wedge(f\vee\neg b)=f\vee\neg b$. Thus $f≤\neg b$, which means that $\neg b\in F'$.
Since $\neg b\in F'$, either $\neg b\in F$ or $\neg b$ may be acquired by meets and upwards closures from $F\cup\{b\}$. Say $b\wedge f≤\neg b$ for some $f\in F$. Then $b\wedge f= b\wedge f\wedge\neg b = b\wedge\neg b\wedge f = 0\wedge f = 0$ for an $f\in F$ and by the above argument, we derive $\neg b\in F$. This is a contradiction, from which we can derive that $0\not\in F$.
Hence, $0\not\in F'$, and thus $F'$ is a proper ideal strictly containing $F$.
| 4 | https://mathoverflow.net/users/102 | 18940 | 12,622 |
https://mathoverflow.net/questions/18928 | 11 | Elementary commutative algebra fact: for two proper ideals I and J of a commutative ring R, we have $V(IJ)=V(I\cap J)=V(I)\cup V(J)$.
Closed subschemes are related to sheaves of ideals. There is operation of intersection and product between sheaves of ideals, which is similar to the affine case.
I see in many places that the structure sheaf over $V(I)\cup V(J)$ are defined to be $R/I\cap J$ rather than $R/IJ$. Why should the structure sheaf be define in that way?
| https://mathoverflow.net/users/2666 | Union of closed subschemes with the structure sheaf over it | Because the first one is the right answer in the case of affine varieties, and the second one is not. Indeed, $R/I$, $R/J$ are nilpotent-free implies $R/I\cap J$ is nilpotent-free, but not so for $R/IJ$.
| 10 | https://mathoverflow.net/users/1784 | 18942 | 12,624 |
https://mathoverflow.net/questions/18939 | 1 | Let $X$ (resp. $Y$) be the affine $k$-scheme defined by the ideal $I$ (resp. $J$) in the polynomial ring $k[x\_1,...x\_n]$ (resp. $k[y\_1,...,y\_m]$).
Let $Z$ be the affine scheme defined by the ideal $L$ in $k[z\_1,...z\_s]$, and let $f^\\*:k[z]/L\rightarrow k[x]/I$ (resp. $g^\\*:k[z]/L\rightarrow k[y]/J$) be $k$-homomorphisms, where $x=(x\_1,...,x\_n)$ and so forth, corresponding to scheme morphisms $f:X\rightarrow Z$ (resp. $Y\rightarrow Z$).
Then it should be possible to express the fiber product $X\times\_{f,Z,g}Y$ via an ideal $W$ in the polinomial ring $k[x,y,z]$ [edit: actually, $W$ should be an ideal in $k[x,y]$] (where $x$ stands for the string of variables $x\_1,...,x\_n$, and so on).
Question: how to express $W\subseteq k[x,y,z]$ explicitely in terms of $I$, $J$, $L$, $f^\\*$ and $g^\\*$?
Edit:
You can express things explicitely in terms of some polynomials $F\_i$, $G\_i$ and $H\_i$ such that $I=(F\_1,...,F\_N)$, $J=(G\_1,...,G\_M)$ and $L=(H\_1,...,H\_S)$, and in terms of the components $(f\_1,...,f\_s)$ (resp. $(g\_1,...,g\_s)$) of $f$ (resp. $g$).
| https://mathoverflow.net/users/4721 | Expressing fiber product of affines via an ideal | Dear unknown, let me first congratulate you on the clearness of your question and the quality of your notation, which I'm now going to use.
The fibre product $X\times\_Z Y$ is the subscheme of $\mathbb A\_k^n \times \mathbb A\_k^m$ described by an ideal $\mathfrak A \subset k[x,y]$.
That ideal is $\mathfrak A=I^e + J^e + D$, where
$I^e$ is the extension of $I\subset k[x]$ to $I^e\subset k[x,y]$,
$J^e$ is the extension of $J\subset k[y]$ to $J^e\subset k[x,y]$,
$D$ is the ideal generated by the $s$ differences $f\_i(x)-g\_i(y),\quad (i=1,\ldots,s)$
I find it clearer not to use generators for $I$ and $J$ and, strangely, $L$ is not used at all: this is because the fibre product is the same whether considered over $Z$ or over
$\mathbb A\_k^s$ !
| 6 | https://mathoverflow.net/users/450 | 18947 | 12,626 |
https://mathoverflow.net/questions/18922 | 6 | Are there books or survey articles explaining the subject to a non-expert? To clarify what I mean, here is a couple of issues that I would like to read about. (I am mainly interested in references but would appreciate answers to these specific questions.)
1) As far as I remember, PA do not have a "built-in" scheme for inductive definitions. So I assume that it is not immediately clear how to define things like $x^n$ or the $n$th Fibonacci number. How do they do things like that? One can define some specific coding of finite sequences of numbers and use that, but this is so ugly and so specific to aritmetics, it there a better way?
2) I vaguely remember that there are arithmetic facts provable in ZF but not in PA. Is this indeed the case? Are there simple explicit examples? Is there a way to understand, at least informally, why PA is not enough? (E.g. a proof may use analysis but why cannot it be reformulated in terms of some kind of "constructible" numbers and functions?)
Background: I am as far from logic as a mathematician can be.
| https://mathoverflow.net/users/4354 | A book explaining power and limitations of Peano Axioms? | For obvious reasons, foundations textbooks (undergraduate or beginning graduate level) tend to have essentially no prerequisites, so I suspect you will find most of them to be accessible.
Here are some higher level recommendations that are more to the point. These assume some rudimentary knowledge of Computability Theory and Model Theory. One well-written book that directly addresses your question is Richard Kaye's *Models of Peano Arithmetic* (Oxford Logic Guides 15, OUP, 1991). It's pretty accessible, but I think it's out of print. Another is *Metamathematics of First-Order Arithmetic* (available on [Project Euclid](http://projecteuclid.org/euclid.pl/1235421927)) by Hájek and Pudlák. This one is still available but I wouldn't recommend it to a beginner. For second-order arithmetic and it's relationship to mainstream mathematics, Simpson's *Subsystems of Second-Order Arithmetic* (2nd ed, CUP, 2010) is the canonical choice. The early parts are very accessible but the later chapters are much denser.
---
Examples for (2) include:
* [Gödel's Incompleteness Theorems](http://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorems)
* [Paris–Harrington Theorem](http://en.wikipedia.org/wiki/Paris%E2%80%93Harrington_theorem)
* [Goodstein's Theorem](http://en.wikipedia.org/wiki/Goodstein%27s_theorem) (mentioned by Aaron Bergman)
* [Kirby–Paris Hydra Games](http://math.andrej.com/2008/02/02/the-hydra-game/)
| 8 | https://mathoverflow.net/users/2000 | 18950 | 12,628 |
https://mathoverflow.net/questions/18847 | 39 | A first course in algebraic topology, at least the ones I'm familiar with, generally gets students to a point where they can calculate homology right away. Building the theory behind it is generally then left for the bulk of the course, in terms of defining singular homology, proof of the harder Eilenberg-Steenrod axioms, cellular chains, and everything else necessary to show that the result is essentially independent of the definitions. A second course then usually takes up the subject of homotopy theory itself, which is harder to learn and often harder to motivate.
This has some disadvantages, e.g. it leaves a discussion of Eilenberg-Maclane spaces and the corresponding study of cohomology operations far in the distance. However, it gets useful machinery directly to people who are consumers of the theory rather than looking to research it long-term.
Many of the more recent references (e.g. tom Dieck's new text) seem to take the point of view that from a strictly logical standpoint a solid foundation in homotopy theory comes first. I've never seen a course taught this way and I'm not really sure if I know anyone who has, but I've often wondered.
So the question is:
*Has anyone taught, or been taught, a graduate course in algebraic topology that studied homotopy theory first? What parts of it have been successful or unsuccessful?*
| https://mathoverflow.net/users/360 | "Homotopy-first" courses in algebraic topology | I was a heavily involved TA for such a graduate course in 2006 at UC Berkeley.
We started with a little bit of point-set topology introducing the category of compactly generated spaces. Then we moved into homotopy theory proper. We covered CW-complexes and all the fundamental groups, Van-Kampen's Theorem, etc. From this you can prove some nice classical theorems, like the Fundamental Theorem of Algebra, the Brauwer Fixed Point Theorem, the Borsuk-Ulam Theorem, and that $R^n \neq R^m$ for $n \neq m$. I felt like this part of the course went fairly well and is sufficiently geometric to be suitable for a first level graduate course (you can draw lots of pictures!).
At this point you can take the course in a couple different directions which all seem to have their own disadvantages and problems. The main problem is lack of time. A very natural direction is to discuss obstruction theory, since it is based off of the same ideas and constructions covered so far. However this is not really possible since the students haven't seen homology or cohomology at this point!
Instead, for a bit we discussed the long exact sequences you get from fibrations and cofibrations. You could then try to lead into the definition of cohomology as homotopy classes of maps into a $K(A,n)$. But this definition is fairly abstract and doesn't show one of the main feature of homology/cohomology: It is extremely computable. Still, I could imagine a course trying to develop homology and cohomology from this point of view and leading into CW homology and the Eilenberg-Steenrod axioms.
Another direction you can go is into the theory of fiber bundles (this is what we tried). The part on covering space theory works fairly well and you have all the tools at your disposal. However when you want to do general fiber bundle theory it can be difficult. A natural goal is the construction of classifying spaces and Brown's representability theorem.
The problem is that the homotopy invariance of fiber bundles is non-trivial to prove. You should expect to have to spend fair amount of time on this. It is really more suited for a second course on algebraic topology.
The main problem with all of these approaches is that it is difficult to cover the homotopy theory section and still have enough time to cover homology/cohomology properly. You know this has to be the case since it is hard to do the reverse: cover homology and cohomology, and still have enough time to cover homotopy theory properly.
What this means is that you'll be in the slightly distasteful situation of having bunch of students who have taken a first course on algebraic topology, but don't really know about homology or cohomology. This is fine if you know that these students will be taking a **second** semester of algebraic topology. Then any gaps can be fixed. However, in my experience this is not a realistic expectation. As you well know, you will typically have some students who end up not being interested in algebraic topology and go into analysis or algebraic geometry or some such. Or you might have some students who are second or third year students in other math fields and are taking your course to learn more about homology and cohomology. They would be done a particular disservice by a course focusing on "homotopy first".
| 30 | https://mathoverflow.net/users/184 | 18955 | 12,631 |
https://mathoverflow.net/questions/18957 | 7 | SEEKING REFERENCES FOR SARRUS' RULE AND EXTENSIONS
An undergraduate came to me with an identity for 4x4 determinants that is actually correct:
$\det(A)=h(A)+h(RA)+h(R^{2}A)$
where R cyclically permutes the last three rows of the matrix A. I wont define h here but, except for the signs of the terms, it is the usual incorrect extension of Sarrus Rule that is familiar to anyone who has has taught linear algebra (His identity is not the Laplace expansion, as it has three 4x4 matrices, rather than four 3x3 matrices. )
Is something like this known? Mathscinet lists a paper Monaco and Monaco that might be relevant, but my library couldnt get it. I haven't even found the original reference for Sarrus' rule itself.
Eric Schmutz
| https://mathoverflow.net/users/4796 | Sarrus determinant rule: references, extensions | I have found references to the Sarrus determinant rule and to an extension of it. In The Quarterly journal of pure and applied mathematics, Volume 38 which is available on Google books in the article "A fourth list of writings on determinants". On page 239 There is a reference to what I believe is Sarrus original result. Then on page 350 there is a reference to an extension of Sarrus' original result. I believe that part one of Muir's book on determinants and their history is available on Google books
| 1 | https://mathoverflow.net/users/1098 | 18961 | 12,634 |
https://mathoverflow.net/questions/18959 | 12 | This is a followup to [Analog to the Chinese Remainder Theorem in groups other than Z\_n.](https://mathoverflow.net/questions/18893/analog-to-the-chinese-remainder-theorem-in-groups-other-than-z-n) . I shouldn't have used the comments to ask a new question, in fact...
Here is the statement of the Chinese Remainder Theorem, as it occurs in most books and websites:
(1) Let $R$ be a commutative ring with unity, and $I\_1$, $I\_2$, ..., $I\_n$ be finitely many ideals of $R$ such that ($I\_i+I\_j=R$ for any two distinct elements $i$ and $j$ of $\left\lbrace 1,2,...,n\right\rbrace$). Then, $I\_1I\_2...I\_n=I\_1\cap I\_2\cap ...\cap I\_n$, and the canonical ring homomorphism $R/\left(I\_1I\_2...I\_n\right)\to R/I\_1 \times R/I\_2 \times ... \times R/I\_n$ is an isomorphism.
But there seems to be another, even more general form of (1) which doesn't get even half of the attention:
(2) Let $R$ be a commutative ring with unity, and $I\_1$, $I\_2$, ..., $I\_n$ be finitely many ideals of $R$ such that ($I\_i+I\_j=R$ for any two distinct elements $i$ and $j$ of $\left\lbrace 1,2,...,n\right\rbrace$). Let $A$ be an $R$-module. Then, $I\_1I\_2...I\_n\cdot A=I\_1A\cap I\_2A\cap ...\cap I\_nA$, and the canonical $R$-module homomorphism $A/\left(I\_1I\_2...I\_n\cdot A\right)\to A/I\_1A \times A/I\_2A \times ... \times A/I\_nA$ is an isomorphism.
I am wondering: is (2) a trivial corollary of (1)? Because otherwise I don't see any reason why (2) shouldn't appear in literature as "the" Chinese Remainder Theorem, with (1) being but a corollary. Or is (2) wrong? The only way I see to get (2) from (1) is to apply (1) to the ring $R\oplus A$ (with multiplication on $R\oplus 0$ inherited from $R$, multiplication between $R\oplus 0$ and $0\oplus A$ given by the $R$-module structure on $A$, and multiplication on $0\oplus A$ given by $0$), which seems quite artificial to me. Am I missing something very obvious?
| https://mathoverflow.net/users/2530 | Chinese Remainder Theorem for rings: why not for modules? | The second result you're talking about is also sometimes called the Chinese remainder theorem, and can be derived from the Chinese remainder theorem for rings by "tensoring the CRT isomorphism" with $A$. Explicitly, (1) gives
$R/\prod\_{k=1}^n I\_k\cong\prod\_{k=1}^n R/I\_k$
via the natural map. This is an isomorphism of rings as well as an isomorphism of $R$-modules. Therefore, upon tensoring with $A$, it becomes
$A/\big(\prod\_{k=1}^nI\_k\big)A\simeq \prod\_{k=1}^n A/I\_kA$
via the natural map, using the canonical isomorphism $R/I\otimes\_RA\cong A/IA$ as well as the fact that tensor product commutes with finite direct products. It follows that the kernel of $A\rightarrow\prod\_{k=1}^n A/I\_kA$, which is clearly $\bigcap\_{k=1}^n I\_kA$, is equal to $\big(\prod\_{k=1}^n I\_k\big)A$. So, you've derived (2) from (1). Keep in mind that (2) is an isomorphism of $R$-modules, while (1) is an isomorphism of rings (as well as $R$-modules).
| 16 | https://mathoverflow.net/users/4351 | 18962 | 12,635 |
https://mathoverflow.net/questions/18964 | 17 | Using p-adic analysis, Dwork was the first to prove the rationality of the zeta function of a variety over a finite field. From what I have seen, in algebraic geometry, this method is not used much and Grothendieck's methods are used instead. Is this because it is felt that Dwork's method is not general or powerful enough; for example, Deligne proved the Riemann Hypothesis for these zeta functions with Grothendieck's methods, is it felt that Dwork's method can't do this?
| https://mathoverflow.net/users/4692 | Dwork's use of p-adic analysis in algebraic geometry | The premise of the question is not correct. Dwork's methods (and modern descendants of them) are a major part of modern arithmetic geometry over $p$-adic fields, and of $p$-adic analysis.
You could look at the two volume collection *Geometric aspects of Dwork theory* to get
some sense of these developments.
Just to say a little bit here: Dwork's approach led to Monsky--Washnitzer cohomology,
which in turn was combined with the theory of crystalline cohomology by Bertheolt to develop the modern theory of rigid cohomology. The $p$-adic analysis of Frobenius actions is also
at the heart of the $p$-adic theory of modular and automorphic forms, and of much of the machinery underlying $p$-adic Hodge theory. The theory of $F$-isocrystals (the $p$-adic analogue of variations of Hodge structure) also grew (at least in part) out of Dworks
ideas.
To get a sense for some of Dwork's techniques, you can look at the Bourbaki report *Travaux de Dwork*, by Nick Katz, and also at Dwork's difficult masterpiece *$p$-adic cycles*, which has been a source of insipiration for many arithmetic geometers.
In some sense the $p$-adic theory is more difficult than the $\ell$-adic theory, which is
why it took longer to develop. (It is like Hodge theory as compared to singular cohomology.
The latter is already a magnificent theory, but the former is more difficult in the sense that it has more elaborate foundations and a more elaborate formalism, and (related to this) has ties to analysis (elliptic operators, differential equations) that are not present
in the same way in the latter.) [For the experts: I am alluding to $p$-adic Hodge theory,
syntomic cohomology, $p$-adic regulators, Serre--Tate theory, and the like.]
| 33 | https://mathoverflow.net/users/2874 | 18966 | 12,637 |
https://mathoverflow.net/questions/18960 | 1 | We have a closed curve C on the plane given by parametric equations: x=x(t), y=y(t), t changes between a and b, x and y are smooth functions.
We want to calculate the winding number of this curve around the origin.
The most natural way to do it is to calculate the path integral:
$$\int\_C \frac{-y\,dx+x\,dy}{x^2+y^2}$$
However, this integral turns out to be too complicated to calculate. What should we do now? Are there any efficient and strong methods to quickly and calculate the winding number?
Thanks.
| https://mathoverflow.net/users/4797 | How To Calculate A Winding Number? | This is simple if you can draw a picture of your curve. Find a direction so that your tangent is always moving as you pass through it. Count the number of tangents pointing in that direction with a sign. +1 if you are moving through the direction counterclockwise, and -1 if you are moving through the direction clockwise. The sum of the +1's and -1's is your winding number.
| 7 | https://mathoverflow.net/users/4304 | 18968 | 12,639 |
https://mathoverflow.net/questions/11713 | 11 | The slice-ribbon conjecture asserts that all slice knots are ribbon.
This assumes the context:
1) A `knot' is a smooth embedding $S^1 \to S^3$. We're thinking of the 3-sphere as the boundary of the 4-ball $S^3 = \partial D^4$.
2) A knot being slice means that it's the boundary of a 2-disc smoothly embedded in $D^4$.
3) A slice disc being ribbon is a more fussy definition -- a slice disc is in ribbon position if the distance function $d(p) = |p|^2$ is Morse on the slice disc and having no local maxima. A slice knot is a ribbon knot if one of its slice discs has a ribbon position.
My question is this. All the above definitions have natural generalizations to links in $S^3$. You can talk about a link being slice if it's the boundary of disjointly embedded discs in $D^4$. Similarly, the above ribbon definition makes sense for slice links. Are there simple examples of $n$-component links with $n \geq 2$ that are slice but not ribbon? Presumably this question has been investigated in the literature, but I haven't come across it. Standard references like Kawauchi don't mention this problem (as far as I can tell).
| https://mathoverflow.net/users/1465 | slice-ribbon for links (surely it's wrong) | Ryan, I think this is an open problem. The best related result I know is a theorem of Casson and Gordon [A loop theorem for duality spaces and fibred ribbon knots. Invent. Math. 74 (1983)] saying that for a fibred knot that bounds a homotopically ribbon disk in the 4-ball, the slice complement is also fibred.
More precisely, they are assuming that the knot K bounds a disk R in the 4-ball such that the inclusion
$S^3 \smallsetminus K \hookrightarrow D^4 \smallsetminus R$
induces an epimorphism on fundamental groups. If one glues R to a fibre of the fibration $S^3 \smallsetminus K \to S^1$ to obtain a closed surface F, then the statement is that the monodromy extends from F to a solid handlebody which is a fibre of a fibration $D^4 \smallsetminus R \to S^1$ extending the given one on the boundary.
| 12 | https://mathoverflow.net/users/4625 | 18971 | 12,641 |
https://mathoverflow.net/questions/18974 | 2 | Let $G$ be a finitely presented group and $N$ a normal subgroup. Is $N$ finitely generated or normally finitely generated? Here normally finitely generation means that for some finite set $S$ of elements in $N$, every element of $N$ can be writen as a product of $G$-conjugation of elements in $S$. Thanks.
| https://mathoverflow.net/users/1546 | Is a normal subgroup of a finitely presented group finitely generated or normal finitely genrated? | If $G$ is the free group on two generators, then $N$ the commutator subgroup is not finitely generated.
If $H$ is any finitely generated, but not finitely presented group, then $H$ is the quotient of a finitely generated free group $G$, with kernel $N$ which is not normally finitely generated.
Steve
| 11 | https://mathoverflow.net/users/1446 | 18975 | 12,644 |
https://mathoverflow.net/questions/18790 | 10 | For every finite dimensional semi-simple Lie group $\mathfrak{g}$, we have a loop algebra $\mathfrak{g}[t,t^{-1}]$. This loop algebra has a natural invariant inner product by taking the residue at zero of the Killing form applied to two elements (i.e. $t^k\mathfrak{g}$ and $t^{-k-1}\mathfrak{g}$ are paired by the Killing form.)
This Lie algebra actually has a Manin triple structure with respect to this inner product: the subalgebras $\mathfrak{g}[t]$ and $t^{-1}\mathfrak{g}[t^{-1}]$ are both isotropic, and non-degenerately paired by this form. This makes $\mathfrak{g}[t]$ into a Lie bialgebra, by getting the cobracket from the bracket on $t^{-1}\mathfrak{g}[t^{-1}]$.
Now, as we all know, Lie bialgebras can be quantized: in this case, the result is a quite popular Hopf algebra called the **Yangian**. By the usual yoga of quantization of Lie bialgebras, the dual Hopf algebra to the Yangian quantizes the universal enveloping $t^{-1}\mathfrak{g}[t^{-1}]$, so if you take a different associated graded of the Yangian, you must get the Hopf algebra of functions on the group with Lie algebra $t^{-1}\mathfrak{g}[t^{-1}]$, which is $L\_<G$, the based formal loop group.
>
> Now, all of these things also have explicit descriptions in terms of equations, and it seems as though this story must be worked out explicitly somewhere, but I've had little luck locating it. Does anyone know where? Or is this story just wrong, and that's why I can't find it?
>
>
>
**EDIT:** The comment below mostly answers this question. I would be interested if anyone out there has written something more explicit than the Etingof and Kazhdan paper, but it's the sort of thing I was looking for. If it were to be left in the form of an answer, I would probably accept it (hint, hint).
| https://mathoverflow.net/users/66 | Is there a good reference for the relationship between the Yangian and formal based loop group? | I will suggest the article [Quantization of Lie bialgebras, III](http://arxiv.org/abs/q-alg/9610030/) by Pavel Etingof and David Kazhdan. It discusses both the Yangian and the dual Yangian as examples in the context of quantization of Lie bialgebras of functions on a curve with punctures.
Also, so far as I know, no-one ever explicitly derives the relations of the Yangian from the conditions that it be a Hopf algebra quantization of the Lie bialgebra, but Drinfeld explains it in his article called Quantum Groups (Proc. ICM Berkeley, 1986). If you assume the coproduct takes a certain form on the Lie algebra (the most obvious choice given that it is a quantization of that Lie bialgebra) then impose the condition that the Hopf algebra coproduct be an algebra homomorphism then you can derive the relations of the Yangian in the first presentation term by term, using the fact that it is a homogeneous (graded) deformation.
Personally I tried to do this but I found I had to assume little things along the way, like that the RHS of equation (13) in that paper is a symmetric sum of the orthonormal basis elements of the Lie algebra. I couldn't really understand why I had to do this, but I guess it's probably obvious to people who are smarter than me. At the end you end up with an algebra that you can then prove is isomorphic to the second presentation he gives, and then get a PBW theorem based on this presentation and verify that it is a quantization of the Lie bialgebra after all (I still working on understanding this part). Then given the uniqueness, which is meant to follow from cohomological arguments described in Section 9 of that same Quantum Groups paper, you don't need to justify any assumptions you make along the way. It seems that understanding the cohomology is key.
| 5 | https://mathoverflow.net/users/3316 | 18983 | 12,648 |
https://mathoverflow.net/questions/18987 | 27 | Using Alexander duality, you can show that the Klein bottle does not embed in $\mathbb{R}^3$. (See for example [Hatcher's book Chapter 3](http://www.math.cornell.edu/~hatcher/AT/ATch3.pdf) page 256.) Is there a more elementary proof, that say could be understood by an undergraduate who doesn't know homology yet?
| https://mathoverflow.net/users/3557 | Why can't the Klein bottle embed in $\mathbb{R}^3$? | If you are willing to assume that the embedded surface $S$ is polyhedral, you can prove that it is orientable by an elementary argument similar to the proof of polygonal Jordan Theorem. Of course the proof is translation of a homology/transversality/separation argument.
Fix a direction (nonzero vector) which is not parallel to any of the faces. For every point $p$ in the complement of $S$, consider the ray starting at $p$ and goint to the chosen direction. If this ray does not intersect edges of $S$, count the number of intersection points of the ray and the surface. If this number is even, you say that $p$ is black, otherwise $p$ is white. If the ray intersects an edge of $S$, you paint $p$ the same color as some nearby point whose ray does not intersect edges. It is easy to see that the color does not change along any path in the complement of $S$ (it suffices to consider only polygonal paths avoiding points whose rays contain vertices of $S$).
Now take points $p$ and $q$ near the surface such that the segment $pq$ is parallel to the chosen direction. Then they are of different colors. But if the surface is non-orientable, you can go from $p$ to $q$ along a Mobius strip contained in the surface. This contradicts the above fact about paths in the complement of $S$.
| 24 | https://mathoverflow.net/users/4354 | 19005 | 12,663 |
https://mathoverflow.net/questions/18982 | 8 | In a letter to Tate from 1987, Serre describes a beautiful Theorem relating mod p modular forms to quaternions ("Two letters on quaternions and modular forms (mod p)", Israel J. Math. 95 (1996), 281--299). At the beginning of Remark (4) on page 284, Serre says that every supersingular elliptic curve E over \bar F\_p has a "canonical and functorial" structure E\_0 over the finite field F with p^2 elements. He further says (if I understand it correctly) that such F-structure is determined by the condition that the relative Frobenius endomorphism of E\_0 is equal to -p in the endomorphism ring of E\_0 over F.
My question is: does anyone have a detailed reference for the proof of this fact, please? Also, can one find an F-structure of a given supersingular elliptic curve E where the relative Frobenius acts as multiplication by p (instead of -p)? Thanks
| https://mathoverflow.net/users/4800 | Supersingular elliptic curves and their "functorial" structure over F_p^2 | (**EDIT:** I've rewritten my argument in terms of the inverse functor, i.e., base extension, since it is clearer and more natural this way.)
Much of what is below is simply a reorganization of what Robin Chapman wrote.
>
> **Theorem:** For each prime $p$, the base extension functor from the category $\mathcal{C}\_{-p}$ of elliptic curves over $\mathbf{F}\_{p^2}$ on which the $p^2$-Frobenius endomorphism acts as $-p$ to the category of supersingular elliptic curves over $\overline{\mathbf{F}}\_p$ is an [equivalence of categories](http://en.wikipedia.org/wiki/Equivalence_of_categories).
>
>
>
**Proof:** To show that the functor is an equivalence of categories, it suffices to show that the functor is full, faithful, and essentially surjective. It is faithful (trivially), and full (because homomorphisms between base extensions of elliptic curves in $\mathcal{C}\_{-p}$ automatically respect the Frobenius on each side). Essential surjectivity follows from Lemma 3.2.1 in
[Baker, González-Jiménez, González, Poonen, "Finiteness theorems for modular curves of genus at least 2", *Amer. J. Math.* **127** (2005), 1325–1387](http://www-math.mit.edu/~poonen/papers/finiteness.pdf),
which is proved by constructing a model for one curve and getting models for the others via separable isogenies. $\square$
---
The same holds for the category $\mathcal{C}\_p$ defined analogously, but with Frobenius acting as $+p$.
Here are two approaches for proving essential surjectivity for $\mathcal{C}\_p$:
1) If $G:=\operatorname{Gal}(\overline{\mathbf{F}}\_p/\mathbf{F}\_{p^2})$ and $E$ is an elliptic curve over $\mathbf{F}\_{p^2}$, and $\overline{E}$ is its base extension to $\overline{\mathbf{F}}\_{p^2}$, then the image of the nontrivial element under $H^1(G,\{\pm 1\}) \to H^1(G,\operatorname{Aut} \overline{E})$ gives the quadratic twist of $E$ (even when $p$ is $2$ or $3$, and even when $j$ is $0$ or $1728$). Applying this to each $E$ with Frobenius $-p$ gives the corresponding elliptic curve with Frobenius $+p$.
2) Use Honda-Tate theory (actually, it goes back to Deuring in this case) to find one supersingular elliptic curve over $\mathbf{F}\_{p^2}$ with Frobenius $+p$, and then repeat the proof of Lemma 3.2.1 to construct the models of all other supersingular elliptic curves via separable isogenies.
| 9 | https://mathoverflow.net/users/2757 | 19013 | 12,668 |
https://mathoverflow.net/questions/19041 | 11 | Is there any possibility of a Poisson Geometry version of the Fukaya category? Given a Poisson manifold Y, objects could be manifolds with isolated singularities X which have the property that TX is contained in NX maximally. The naive example would be something like the Poisson structure on R^2 which is (x^2 + y^2)(d/dx^d/dy). Branes would in this case be curves with some nodal singularity at the origin.
The morphisms could still be from holomorphic disks with respect to the standard complex structure. In principal, it seems like in this example the Fukaya category could be defined in the standard way (although maybe there is something more subtle one should do with the morphisms?). For a brane L passing through the origin there should be some interesting multiplicative structure in the algebra A(L) owing to the fact that the brane is required to remain fixed at the origin.... I would hope that the Hochschild cohomology could be related to the Poisson cohomology of the manifold though I haven't studied my example yet. What obstructions arise when trying to construct this category?
| https://mathoverflow.net/users/6986 | A Poisson Geometry Version of the Fukaya Category | The fundamental technique of symplectic topology is the theory of pseudo-holomorphic curves. One studies maps $u$ from a Riemann surface into a symplectic manifold, equipped with an almost complex structure tamed by the symplectic form, such that $Du$ is complex-linear. Numerous algebraic structures can be built from such maps: Gromov-Witten invariants, Hamiltonian Floer cohomology, Floer cohomology of pairs of Lagrangian submanifolds, and most elaborate of all, $A\_\infty$-structures on Lagrangian Floer cochains (Fukaya categories).
Though the basic theory of pseudo-holomorphic curves makes sense on more general almost-complex manifolds, the presence of the symplectic structure is vital for Gromov compactness to be applicable. Without this, your curves are liable to vanish into thin air. None of the algebraic structures I mentioned have been developed on almost complex manifolds, nor on Poisson manifolds. It's conceivable that leafwise constructions can be made to work in the Poisson context, but there are basic analytic and geometric questions to be addressed.
There are situations where one might reasonably hope to find relations between Poisson geometry and symplectic topology, but in those situations it may be wise to go via intermediate constructions. For instance, a version of the derived Fukaya category of $T^{\ast} L$ was shown by Nadler to be equivalent to the derived category of constructible sheaves on $L$, and I'm told that that category is related to deformation quantization of $T^{\ast} L$ - something which truly does belong to Poisson geometry.
| 6 | https://mathoverflow.net/users/2356 | 19044 | 12,683 |
https://mathoverflow.net/questions/18970 | 8 | When I read descriptions of Apollonius' treatise on conics, some of them say that he invented co-ordinate geometry, some say that he kind of did and others are silent on the matter. Or is it the case that there is no simple answer?
| https://mathoverflow.net/users/4692 | Did Apollonius invent co-ordinate geometry? | Let V be the vertex of a parabola, F its focus, X a point on its symmetry axis, and A a point on the parabola such that AX is orthogonal to VX. It was well within the power of the Greeks to prove relations such as $VX:XA = XA:4VF$. If you introduce coordinate axes, set $x = VX$, $y = XA$ and $p = VF$, you get $y^2 = 4px$, the modern form of the equation of a parabola.
Everything now depends on what "invention of coordinate geometry" means to you. I do not think that the Greeks' work on conics should be confused with coordinate geometry since they did not regard the lengths occurring above as coordinates. It's just that parts of their results are very easily translated into modern language.
In a similar vein, Eudoxos and Archimedes already were close to modern ideas behind integration, but they did not invent calculus. Euclid, despite Heath's claim to the contrary, did not state and prove unique factorization. And Euler, although he knew the product formula for sums of four squares, did not invent quaternions (Blaschke once claimed he did).
In any case, we are much more careful now with sweeping claims such as "Appolonius knew coordinate geometry" than historians were, say, 100 years ago.
| 19 | https://mathoverflow.net/users/3503 | 19050 | 12,685 |
https://mathoverflow.net/questions/19046 | 40 | I want some recommendation on which software I should install on my computer. I'm looking for an open source program for general abstract mathematical purposes (as opposed to applied mathematics).
I would likely use it for group theory, number theory, algebraic geometry and probably polytopes.
The kind of program I have in mind is Mathematica or Matlab. Although probably those are not designed for abstract mathematics.
Any suggestions?
| https://mathoverflow.net/users/4619 | Open source mathematical software | Here are some links.
* [Axiom](http://www.axiom-developer.org/) and [Maxima](http://maxima.sourceforge.net/) are good general purpose computer algebra systems.
* [DataMelt](http://jwork.org/dmelt/) is a free Java-based math software with a lot of examples
* [GAP](http://www.gap-system.org/) is a system for computational discrete algebra (with particular emphasis on computational group theory).
* [PARI/GP](http://pari.math.u-bordeaux.fr/) is a CAS for fast computations in number theory.
* [SAGE](http://www.sagemath.org/) is a kind of unified framework for several systems, including GAP, PARI, and Maxima.
* [Octave](http://www.gnu.org/software/octave/) is a system for numerical computations (it is close to Matlab).
* [Cadabra](http://cadabra.phi-sci.com/) is a computer algebra system designed for the solution of the field theory problems.
* [CoCoA](http://cocoa.dima.unige.it/) stands for "Computations in commutative algebra".
* [KANT / KASH](http://www.math.tu-berlin.de/~kant/kash.html) stands for "Computational Algebraic Number Theory".
* [Macaulay 2](http://www.math.uiuc.edu/Macaulay2/) is a system for research in commutative algebra and algebraic geometry.
* [Snap](http://snap-pari.sourceforge.net/) is a computer program for studying arithmetic invariants of hyperbolic 3-manifolds.
* [Symmetrica](http://www.algorithm.uni-bayreuth.de/en/research/SYMMETRICA/) is an object oriented computer algebra system for representations, combinatorics and applications of symmetric groups.
| 65 | https://mathoverflow.net/users/nan | 19051 | 12,686 |
https://mathoverflow.net/questions/19054 | 2 | I'm looking for a formulas book.
I'm currently student in Communication Systems and we have several courses involving mainly complex analysis, fourier analysis, signal processing, information theory and sometimes other principles and I need a lot of books for all these formulas.
Does someone knows a book with these formulas inside ?
It doesn't need to have demonstration or long explanation, I just need something to replace the stack of books I have.
The *best* one I found so far is the [Gieck](http://rads.stackoverflow.com/amzn/click/0071457747), but there are more than a half of the book I'll never use. I want something more specific for signal processing and information theory.
| https://mathoverflow.net/users/4581 | Complete formulas book for Communication System engineer | I would suggest "Handbook of Formulas and Tables for Signal Processing", by Alexander D. Poularikas.
[This link](http://www.crcnetbase.com/isbn/978-0-8493-8579-7) should help you.
[This](http://rads.stackoverflow.com/amzn/click/0849385792) is the amazon link.
| 5 | https://mathoverflow.net/users/2938 | 19055 | 12,689 |
https://mathoverflow.net/questions/19070 | 15 | I am interested in learning the theory of types, especially in how they can provide a foundation to mathematics different to sets and how they can avoid self-referential paradoxes by stipulating that a collection of objects of type n has type n+1. As for my background knowledge, I only know a little of propositional and predicate logic and Zermelo-Fraenkel set theory.
| https://mathoverflow.net/users/4692 | Reference request for type theory | I would suggest you look at Martin-Löf's work, such as the following reprint of his earlier unpublished manuscript (from 1972?):
* Per Martin-L: [An Intuitionistic Theory of Types](http://books.google.si/books?id=pLnKggT_In4C&printsec=frontcover&hl=en&source=gbs_v2_summary_r&cad=0#v=onepage&q=&f=false). In: Twenty-Five Years of Constructive Type Theory Proceedings of a Congress held in Venice, October 1995. Editors: Giovanni Sambin and Jan M. Smith. Oxford University Press, 1998.
I belive a fairly good approximation of this paper is [available online](http://www.cs.chalmers.se/~peterd/kurser/tt03/martinlof72.ps). If you are looking for other online references, have a look at [this lecture by Martin-Löf](http://www.hf.uio.no/ifikk/filosofi/njpl/vol1no1/meaning/meaning.html).
This should give you some idea for type theory as foundation of mathematics.
| 14 | https://mathoverflow.net/users/1176 | 19071 | 12,701 |
https://mathoverflow.net/questions/19066 | 12 | Is there a standard notion of non-degeneracy for multilinear forms?
My motivation is simple curiosity, by the way!
| https://mathoverflow.net/users/1409 | Non-degenerate multilinear forms | I'm not sure if this notion is "standard", but there is one such notion, used for example in Nigel Hitchin's paper [*Stable forms and special metrics*](http://www.ams.org/mathscinet-getitem?mr=1871001) ([arXiv:math/0107101](http://arxiv.org/abs/math/0107101)) for alternating multilinear forms. The idea is that symplectic structures on a vector space $V$ can be characterised by the fact that they lie in an open orbit of $\mathrm{GL}(V)$ on $\Lambda^2V^\*$. Hitchin calls these *stable forms* and shows that apart from the the case of symplectic forms, there are stable $3$-forms in dimensions 6,7 and 8; the $G\_2$-invariant $3$-form in a seven-dimensional vector space (i.e., the imaginary component of the multiplication of imaginary octonions) being one such example.
Hitchin's notion is very fruitful, as it provides a variational approach to 7-dimensional riemannian metrics of weak $G\_2$ holonomy, for example.
| 10 | https://mathoverflow.net/users/394 | 19073 | 12,702 |
https://mathoverflow.net/questions/19032 | 4 | Let E be a Hilbert C\*-module over some C\*-algebra and let $h \in K(E)$. Due to B. Blackadar's, "K-Theory for Operator algebras" Thm. 17.11.4 for a separable C\*-algebra $A$, represented by elements of $B(E)$, it is possible to construct a countable approximate unit $\{u\_n\}$ contained in $C^\*(h)$, such that $\{u\_n\}$ is quasicentral for $A$ and $u\_{n+1} u\_n=u\_n$. The question is: is it always possible to make $u\_n$ be projectors (or, at least, idempotents). The question seems to be obvious if $E$ is just a Hilbert space, but I'm not sure for Hilbert modules.
| https://mathoverflow.net/users/4807 | Approximate unit for the algebra C*(h) consisting of projectors | No. In fact, K(E) need not even contain any nonzero projections. Take a (nontrivial) C\*-algebra B with no nonzero projections1 and take E=B as a right module over B with inner product 〈a,b〉=a\*b. Then K(E)≅B, as mentioned in [Example 13.2.4 (a)](http://books.google.com/books?id=_YQvFHnD6bQC&lpg=PA109&ots=w9jB2qq8x4&dq=blackadar%252013.2.4&pg=PA109#v=onepage&q=&f=true) in Blackadar, and proved for instance as [Proposition 2.2.2 (i)](http://books.google.com/books?id=s_pNcoqXwFoC&lpg=PA203&dq=manuilov%2520troitsky&lr=&pg=PA19#v=onepage&q=&f=true) in Manuilov and Troitsky. (Note that in this case K(E) [also has no nonzero idempotents](https://mathoverflow.net/questions/16943/reference-needed-for-every-idempotent-in-a-c-algebra-is-similar-to-a-hermitian).)
**Edit:** I removed an overly complicated comment on the Hilbert space case, forgetting to take into account that h is strictly positive, and in particular positive. I added a comment on trouble that may arise even in this case.
In the case when E is a Hilbert space over B=ℂ, taking into account the fact that h is self-adjoint, C\*(h) is the C\*-algebra generated by a self-adjoint compact operator, and therefore the spectral projections of h provide an approximate identity {un} consisting of increasing projections. Because h is *strictly* positive, its range is dense, so this will be a sequence of projections converging weakly to the identity operator. However, this approximate identity need not be quasicentral for A⊆B(E). E.g., suppose you have un equal to the projection onto the span of the first n elements of an orthonormal basis. If S is the unilateral shift with respect to that basis, then ||unS−Sun|| = 1 for all n, so {un} is not quasicentral for C\*(S). Pedersen uses the Hahn-Banach theorem and the axiom of choice to show the existence of a quasi-central approximate identity in the closed convex hull of {un}, but you typically will not be able to find one consisting of projections, even in the Hilbert space case.
1 E.g., take B to be C0(X) for some noncompact, locally compact, connected space X, or if you prefer simple algebras see [Blackadar's 1980 paper](http://www.jstor.org/stable/2042420).
| 3 | https://mathoverflow.net/users/1119 | 19075 | 12,703 |
https://mathoverflow.net/questions/19076 | 33 | Some MOers have been skeptic whether something like *natural number graphs* can be defined coherently such that every finite graph is isomorphic to such a graph. (See my previous questions [[1](https://mathoverflow.net/questions/17989/can-every-finite-graph-be-represented-by-one-prescribed-sequence-of-natural-numbe)], [[2](https://mathoverflow.net/questions/17875/can-every-finite-graph-be-represented-by-an-arithmetic-sequence-of-natural-number)], [[3](https://mathoverflow.net/questions/18562/uniformly-computable-classes-of-graphs)], [[4](https://mathoverflow.net/questions/11647/natural-models-of-graphs)])
Without attempting to give a general definition of *natural number graphs*, I invite you to consider the following
>
> **DEFINITION**
>
>
> A natural number $d$ may be called **demi-prime**
> iff there is a prime number $p$ such
> that $d = (p+1)/2$. The demi-primes' distribution is exactly
> like the primes, only shrinked by the
> factor $2$:
>
>
> $$2, 3, 4, 6, 7, 9, 10, 12, 15, 16, 19, 21, 22, 24, 27, 30, 31, 34, 36, 37, 40, 42, 45, 49, ...$$
>
>
> Let **D**($k,n$) be the set which consists
> of the $k$-th up to the $(k+n-1)$-th
> demi-prime number.
>
>
>
After some - mildly exhaustive - calculations I feel quite confident to make the following
>
> **CONJECTURE**
>
>
> For every finite graph $G$ there is a $k$ and a bijection $d$ from the vertex set
> $V(G)$ to **D**($k,|G|$) such that $x,y$ are adjacent if and only if $d(x),d(y)$ are **coprime**.
>
>
>
I managed to show this rigorously for all graphs of order $n\leq $ 5 by brut force calculation, having to take into account all (demi-)primes $d$ up to the 1,265,487th one for graphs of order 5. For graphs of order 4, the first 1,233 primes did suffice, for graphs of order 3 the first 18 ones.
Looking at some generated statistics for $n \leq$ 9 reveals interesting facts(1)(2), correlations, and lack of correlations, and let it seem probable (at least to me) that the above conjecture also holds for graphs of order $n >$ 5.
Having boiled down my initial intuition to a concrete predicate, I would like to pose the following
>
> **QUESTION**
>
>
> Has anyone a clue how to prove or disprove the above conjecture?
>
>
>
My impression is that the question is about the randomness of prime numbers: Are they distributed and their corresponding demi-primes composed randomly enough to mimick – via **D**($k,n$) and coprimeness – all (random) graphs?
---
(1) E.g., there is one graph of order 5 - quite unimpressive in graph theoretic terms - that is very hard to find compared to all the others: it takes 1,265,487 primes to find this guy, opposed to only 21,239 primes for the second hardest one. (Lesson learned: Never stop searching too early!) It's – to whom it is of interest – $K\_2 \cup K\_3$:
```
0 1 0 0 0
1 0 0 0 0
0 0 0 1 1
0 0 1 0 1
0 0 1 1 0
```
(2) **Added:** This table shows the position of the smallest prime (among all primes) needed to mimick the named graphs of order $n$. All values not shown are greater than $\approx 2,000,000$
```
order | 3 4 5 6 7 8
-------------------------------------------------
empty | 14 45 89 89 89 3874
complete | 5 64 336 1040 10864 96515
path | 1 6 3063 21814
cycle | 5 112 21235 49957
```
| https://mathoverflow.net/users/2672 | Bringing Number and Graph Theory Together: A Conjecture on Prime Numbers | **Theorem:** Schinzel's hypothesis H implies the conjecture.
**Proof:** Choose distinct primes $q\_S > 100|G|$ indexed by the 2-element subsets $S$ of $G$. For each $i \in G$, let $Q\_i$ be the set of $q\_S$ for $S$ such that $i \in S$ and the edge $S$ is not part of $G$. Let $P\_i$ be the product of the primes in $Q\_i$. Let $P = 4 \prod\_S q\_S^2$.
By the Chinese remainder theorem, for each $i$ we can find a positive integer $a\_i$ such that
$a\_i \equiv 1 \bmod{\ell^2}$ for each prime $\ell \le 10|G|$,
$a\_i \equiv q-1 \bmod{q^2}$ for each $q \in Q\_i$, and
$a\_i \equiv 1 \bmod{q\_S}$ for each $q\_S \notin Q\_i$.
Moreover, we can choose the $a\_i$ to be distinct. Let $J$ be the set of positive integers up to $\operatorname{max} a\_i$, but excluding all of the $a$'s themselves (i.e., $J$ consists of the numbers in the gaps). For each $j \in J$ choose a prime $s\_j$ much larger than all the $a\_i$ and all the $q\_S$.
Consider the linear polynomials $P n + a\_i$ and $(P n + a\_i + 1)/(2P\_i)$ In $\mathbf{Z}[n]$. For each prime $\ell \le 10|G|$ and each $\ell$ of the form $q\_S$, all these $2|G|$ polynomials are nonzero mod $\ell$ at $n=0$. For each other prime $\ell$, there exists $n$ such that all these polynomials are nonzero mod $\ell$, since $n$ needs to avoid no more than $2|G|$ residue classes mod $\ell$. Furthermore, we can impose the condition that $P n+j$ is divisible by $s\_j^2$ for each $j \in J$, and still find $n$ as above. Therefore Schinzel's hypothesis H implies that there exist arbitrarily large positive integers $n$ such that the numbers $P n+a\_i$ and $(P n + a\_i + 1)/(2P\_i)$ are all prime, and such that $P n+j$ is not prime for $j \in J$. This makes the numbers $p\_i:=P n + a\_i$ *consecutive* primes such that $(p\_i+1)/2 = P\_i r\_i$ for some prime $r\_i$. If $n$ is sufficiently large, then these primes $r\_i$ are all distinct and larger than all of the $q\_S$. So the greatest common factor of $(p\_i+1)/2$ and $(p\_j+1)/2$ for $i \ne j$ equals $1$ if there is an edge between $i$ and $j$, and $q\_{\{i,j\}}$ otherwise. $\square$
---
**Remark:** Given how little is known about consecutive primes, it seems unlikely that the conjecture can be proved unconditionally. But at least now we can be confident that it's true!
| 40 | https://mathoverflow.net/users/2757 | 19080 | 12,706 |
https://mathoverflow.net/questions/17010 | 15 | I am wondering if there is a more "geometric" formulation of complex orientations for cohomology theories than just a computation of $E^\*\mathbb{C}$P$^{\infty}$ or a statement about Thom classes. It seems that later in Hopkins notes he says that the complex orientations of E are in one to one correspondence with multiplicative maps $MU \rightarrow E$, is there a treatment that starts with this perspective? How do the complex orientations of a spectrum E help one compute the homotopy of $E$, or the $E$-(co)homology of MU? Further, what other kinds of orientations could we think about, are there interesting $ko$ or $KO$ orienations? how much of these $E$-orientations of X is detected by E-cohomology of X?
I do have some of the key references already in my library, for example the notes of Hopkins from '99, Rezk's 512 notes, Ravenel, and Lurie's recent course notes. If there are other references that would be great. I am secretly hoping to get some insight from some of the experts. (I guess I should really also go through Tyler's abelian varieties paper)
(sorry for the on and off texing but the preview is giving me weird feedback.)
EDIT: I eventually found the type of answer i was looking for in some notes of Mark Behrens on a course he taught. This answer is that a ring spectrum $R$ is complex orientable is there is a map of ring spectra $MU \to R$. This also appears in COCTALOS by Hopkins but neither source takes this as the more fundamental concept. Anyway, the below answer is more interesting geometrically.
| https://mathoverflow.net/users/3901 | Complex orientations on homotopy | The natural starting point of this story are E-orientations on, say closed, manifolds M. That's just a *fundamental class* $[M^n] \in E\_n(M)$ such that cap product induces a (Poincare duality) isomorphism.
Given E, the question becomes which M are E-orientable. In many cases it happens that this follows if the stable normal bundle of M admits a lift through a fibration $X\to BO$. For example, if E=HZ is ordinary Z-cohomology then X=BSO works, if E=KO then X=BSpin works, if E=KU then X=BU or X=BSpin$^c$ works etc.
To formalize the idea that every X-manifold has an E-orientation, form the bordism groups $\Omega^X\_n$ of X-manifolds and observe that the fundamental classes lead to natural maps $\Omega^X\_n(Y) \to E\_n(Y)$ for any space Y. In other words, there are natural transformations of cohomology theories $\Omega^X \to E$, or even better, maps of spectra $MX \to E$, where $MX$ is the Thom spectrum associated to the fibration $X\to BO$.
In the case X=BU this is called a complex orientation of E and has been studied extensively because it simplifies computations of E-cohomology tremendously. The original and still relevant reference is Adams' little blue book.
| 21 | https://mathoverflow.net/users/4625 | 19085 | 12,708 |
https://mathoverflow.net/questions/19084 | 1 | If $\mathcal M$ is a model category and I know that $A$ and $B$ are isomorphic in $\mathrm{Ho}(\mathcal M)$, is it guaranteed that there is a zig-zag of weak-equivalences in $\mathcal M$ connecting $A$ and $B$?
| https://mathoverflow.net/users/4466 | Equivalences in Model Categories | Yes. The isomorphism in $\mathrm{Ho}(\mathcal{M})$ is represented by a morphism in $\mathcal{M}$ from a cofibrant replacement for $A$ to a fibrant replacement for $B$. The "converse to the Whitehead lemma" states that a map in a model category is a weak equivalence iff its image in the homotopy category is an isomorphism. Combining this with the definition of (co)fibrant replacement, we see that $A$ and $B$ are connected by a 3-step zig-zag of weak equivalences.
| 6 | https://mathoverflow.net/users/126667 | 19089 | 12,711 |
https://mathoverflow.net/questions/19091 | 1 | Let $X$ be a complex algebraic variety, and let $F=(F\_n)$ be a lisse $\mathbb Z\_{\ell}$-sheaf on $X.$ Does there exist an analytic open covering of $X(\mathbb C),$ such that $F$ is "(locally) constant" on each open subset?
| https://mathoverflow.net/users/370 | lisse sheaf on complex varieties | Dear Shenghao, If you really do mean a lisse sheaf on the etale site of $X$, then it doesn't make sense a priori to evaluate it on analytic open subsets of $X(\mathbb C)$, since these are not in the etale site of the algebraic variety $X$. However, $F$ corresponds to a representation of the (profinite) etale $\pi\_1$ of $X$, which in turn is the profinite completion of the
topological $\pi\_1$ of $X(\mathbb C)$. So there is a corresponding locally constant (in the analytic topology) $\mathbb Z\_{\ell}$
sheaf on $X(\mathbb C)$, which, being locally constant, will be constant on sufficiently small analytic open subsets.
| 7 | https://mathoverflow.net/users/2874 | 19093 | 12,712 |
https://mathoverflow.net/questions/19008 | 6 | My problem is perhaps a general lack of understanding but it occurred in a special case of a theorem in Eisenbud's and Harris' "The geometry of schemes" (Theorem VI-29). Let $K$ be a field and $n\in\mathbb{N}$.
>
> Given a closed subscheme $X$ of $\mathbb{P}^n$ with Hilbert Polynomial $P$. The tangent space $T\_{[X]}$ to the Hilbert scheme $\mathcal{H}\_{P}$ at the point corresponding to $X$ is the space of global sections of the normal sheaf $\mathcal{N}\_{\mathbb{P}^n/X}$.
>
>
>
I don't know what "is" precisely means here. $T\_{[X]}$ has a natural $K([X])$-vector space structure and I would like to have an isomorphism of $K([X])$-vector spaces but my problem begins much earlier in the definition of $\mathcal{H}om$-sheaves:
Let $J$ be the ideal sheaf of $X$. The presheaf
$$
U \mapsto \mathcal{H}om\_{\mathcal{O}\_X|\_U}(J/J^2|\_U,\mathcal{O}\_X|\_U)
$$
is already a sheaf of $\mathcal{O}\_X$-modules (this does not depend on the special arguments here). This sheaf $\mathcal{N}\_{\mathbb{P}^n/X}$ is called the normal sheaf of $X$ in $\mathbb{P}^n$. The global sections
$$
\mathcal{N}\_{\mathbb{P}^n/X}(X)=\mathcal{H}om\_{\mathcal{O}\_X}(J/J^2|\_X,\mathcal{O}\_X)
$$
of this normal sheaf is an $\mathcal{O}\_X(X)$-module. Is there an isomorphism of $\mathcal{N}\_{\mathbb{P}^n/X}(X)$ to some $hom$ of rings or modules where I can work with? What is a possible $K([X])$-vector space structure on it?
| https://mathoverflow.net/users/2625 | Question on a theorem of Eisenbud's and Harris' "The geometry of schemes" | Dear roger123,
This is largely a response to your question aksed as a comment below Charles Siegel's answer, but it won't fit in the comment box. Since $\mathcal N\_{\mathbb P^n/X}$ is a sheaf of modules over the sheaf of rings $\mathcal O\_{\mathbb P^n}$ (the structure sheaf of projective space),
its global sections $\mathcal N\_{\mathbb P^n/X}(\mathbb P^n)$ are a module over the ring $\mathcal O\_{\mathbb P^n}(\mathbb P^n)$, which in turn are just $k$ (the ground field).
In short, the global sections of $\mathcal N\_{\mathbb P^n/X}$ form a $k$-vector space.
Maybe you are being confused by the fact that $\mathcal N\_{\mathbb P^n/X}$ is a sheaf on $\mathbb P^n$ that is supported on $X$, so that people often simultaneously regard it as a sheaf on either $X$ or $\mathbb P^n$. This is okay, because if $U$ is any open in $\mathbb P^n$ and $\mathcal F$ is a sheaf supported on $X$, then the sections over an open subset $U$ of $\mathbb P^n$
(when it is regarded as a sheaf on $\mathbb P^n$) will coincide with the sections over
$U\cap X$ (when it is regarded as a sheaf on $X$).
In particular, one has the equation $\mathcal N\_{\mathbb P^n/X}(X) =
\mathcal N\_{\mathbb P^n/X}(\mathbb P^n)$ (an abuse of notation if taken literally; however
one is supposed to regard $\mathcal N\_{\mathbb P^n/X}$ as a sheaf on $X$ on the left-hand side,
and as a sheaf on $\mathbb P^n$ on the right-hand side).
One more thing: If the residue field of the Hilbert scheme at the point $P$ is $k(P)$,
then $P$ is a $k(P)$-valued point of the Hilbert scheme,
and so the corresponding closed subscheme $X$ lies in $\mathbb P^n\_{k(P)}$. Thus, in the
above discussion, $k$ can (and should) be taken to be $k(P)$. Thus the above discussion explains
why $\mathcal N\_{\mathbb P^n/X}(X)$ is a $k(P)$-vector space, as it should be.
| 4 | https://mathoverflow.net/users/2874 | 19094 | 12,713 |
https://mathoverflow.net/questions/19092 | 8 | I wonder whether any of you guys has already read the homonymous note by R. Beals in the December 2009 issue of the **Monthly**.
If so, **would you be so kind as to let me know about the main ideas in Beal's approach**? As you know, the whole point of his note is to present a solution to the following exercise in Herstein's *Topics in Algebra*:
Let **G** be an abelian group having subgroups of order *m* and *n*. Prove that **G** also possesses a subgroup of order **lcm**(*m*, *n*).
The funny thing about this proposal is that in subsequent editions of his book, Prof. Yitz would proclaim that he himself didn't have a solution using the *authorized* tools. Besides, he even went on to saying: "I've had more correspondence about this problem than about any other point in the whole book.".
Being aware of some of the history behind this little pearl, I'd really like to know what it is that Beals came up with. Is his approach crystal-clear? Is it somehow related to the standard attack of proving it first for the case **gcd**(*m*, *n*)=1?
Thanks in advance for you insightful replies.
P.S. The local library is the only access that I have to the literature. Unfortunately, they don't subscribe to any of the MAA periodicals.
1 Beals, Robert. "On Orders of Subgroups in Abelian Groups: An Elementary Solution of an Exercise of Herstein." The American Mathematical Monthly, Vol. 116, No. 10 (Dec., 2009), pp. 923-926; <https://maa.tandfonline.com/doi/abs/10.4169/000298909X477032> <https://www.jstor.org/stable/40391251>
| https://mathoverflow.net/users/1593 | On order of subgroups in abelian groups | The question is to prove that if $H$ and $K$ are subgroups of a
finite Abelian group or orders $m$ and $n$ then $G$ has a subgroup of order
$\mathrm{lcm}(m,n)$.
Beals starts by doing the case where $H$ and $K$ are cyclic. He proves
that $H$ is an internal direct product of cyclic groups of prime power orders.
Then he proves that a product of cyclic subgroups of coprime orders is
cyclic of the right order. The cyclic case is proved by breaking
up $H$ and $K$ as products of cyclic prime power groups, taking
the larger one for each prime and multiplying them up.
The general case follows roughly the same line. Proving that a product
of subgroups of coprime orders has the right order is straightforward.
But decomposing a subgroup into prime power factors using
results earlier in Herstein is more involved. Beals uses Theorem 2.5.1
in Herstein that $|HK| = ~|H||K|/|H\cap K|$. Then Beals finishes
the proof in the same way as the cyclic case.
| 9 | https://mathoverflow.net/users/4213 | 19096 | 12,715 |
https://mathoverflow.net/questions/19090 | 6 | The sine and cosine rules for triangles in Euclidean, spherical and hyperbolic spaces can be understood as invariants for triples of lines. These invariants are given in terms of the distance (both lengthwise and angular) between pairs of lines. There is also a converse statement. Suppose we are living in a complete Riemannian manifold of constant curvature. If a certain sine/cosine rule is satsifies by triples of lines, then we can determine the curvature.
I'm wondering if we can generalize these sine and cosine rules to arbitrary symmetric spaces. That is, give a triple of geodesics (or parallel submanifolds, if we consider higher dimensions), are there similar invariants? Perhaps these invariants will be given in terms of representations of the coset of symmetries take a geodesic to antoher.
It would also be great if these invariants can characterize the symmetric space we are living in, just like in the case of the constant curvature case.
| https://mathoverflow.net/users/nan | Generalizing cosine rule to symmetric spaces | There exist generalizations of the trigonometry laws to symmetric spaces. The following [article](http://arxiv.org/PS_cache/math-ph/pdf/9910/9910041v1.pdf) by Ortega and Santander works out the trigonometry laws for the case of real symmetric spaces of constant curvature and the following [one](http://arxiv.org/abs/math-ph/0112030) the case of rank-1 Hermitian symmetric spaces. The following [article](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.7.9825&rep=rep1&type=pdf.) by Aslaksen and Huynh treats more general cases of n-point trigonometrey of symmetric spaces.
| 6 | https://mathoverflow.net/users/1059 | 19099 | 12,717 |
https://mathoverflow.net/questions/19074 | 5 | Let J be a closed interval of real numbers whose length is finite and positive. Let f be a real valued
function defined on J which has a continuous second derivative at all points of J.
QUESTION: If P1,P2,P3 are any three pairwise distinct and non-collinear points on the graph of f(J), does
there always exist at least one point p on J such that the absolute value of the curvature of this graph
at the point (p,f(p)) is greater than or equal to 1/r-where r is the radius of the circle through the three
points P1,P2,P3?
| https://mathoverflow.net/users/4423 | Can one "soup-up" the LAW OF THE MEAN in the following way? | The answer is yes. Suppose the contrary and rescale the picture so that $r=1$. We may assume that $P\_1$ and $P\_3$ are the endpoints of the graph. There must be points on the graph that are outside the circle - otherwise the curvature at $P\_2$ is at least 1. WLOG assume that there are points below the circle. The lower half of the circle is the graph of a function $f\_0$ defined on an interval of length 2. Let $q$ be a point where $f\_0-f$ attains its maximum, then $f'(q)=f\_0'(q)$. We may assume that $f'(q)=f\_0'(q)\ge 0$, otherwise reflect the picture through the $y$-axis.
I claim that $f(t)<f\_0(t)$ for all $t\ge q$ such that both $f(t)$ and $f\_0(t)$ are defined, contrary to the fact that $P\_2$ lies on the circle. To prove this, consider arc-length parametrizations $\gamma(t)=(x(t),y(t))$ and $\gamma\_0(t)=(x\_0(t),y\_0(t))$ of the two graphs, where $\gamma(0)=(q,f(q))$ and $\gamma\_0(0)=(q,f\_0(q))$. Then
$$
\gamma'(0)=\gamma\_0'(0)=(\cos\alpha,\sin\alpha)
$$
where $0\le\alpha<\pi/2$. Since the curvature of $\gamma$ is less than 1, we have
$$
\angle(\gamma'(t),\gamma'(0)) < t
$$
for all $t>0$. Therefore $x'(t)>\cos(\alpha+t)$ and $y'(t)<\sin(\alpha+t)$ for $0<t<\pi/2-\alpha$. And for $\gamma\_0$ these inequalities turn to equalities. The integration yields that $x(t)\ge x\_0(t)$ and $y(t)< y\_0(t)$ for all $t\in[0,\pi/2-\alpha]$. Since $f\_0$ is increasing after $q$, these inequalities imply that $\gamma(t)$ is below the half-circle (or has already left the domain where $f\_0$ is defined).
Since $\gamma\_0(\pi/2-\alpha)$ is the rightmost point of the circle, the inequality $x(t)\ge x\_0(t)$ for $t=\pi/2-\alpha$ means that the $x$-coordinate of $\gamma(\pi/2-\alpha)$ is already outside the domain of $f\_0$ and one does not need to care about larger values of $t$.
| 4 | https://mathoverflow.net/users/4354 | 19101 | 12,718 |
https://mathoverflow.net/questions/19040 | 3 | This question is basically on applying the Grothendieck-Riemann-Roch theorem to finding a formula for the push-forward of a line bundle on $\mathbf{P}^r$ under a certain morphism. Since I have a lot of questions, let me begin with an example.
Let $\pi:\mathbf{P}^1 \longrightarrow \mathbf{P}^1$ be the morphism defined as $[x\_0:x\_1] \mapsto [x\_0^n:x\_1^n]$. Here $\mathbf{P}^1$ denotes the projective line over $\mathbf{C}$ and $n\geq 1$ is an integer. Note that $\pi$ is finite. For, it is locally given by the map $x\mapsto x^n$.
Let $X=\mathbf{P}^1$.
One can show that $\pi\_\ast (\mathcal{O}(m))$ is given (up to isomorphism) by $$ \mathcal{O}(n(m+1)-1)\oplus \ldots \oplus \mathcal{O}(n(m+1)-1).$$ Here the sum is taken $n$ times. For example, $\pi\_\ast \mathcal{O}\_X \cong \mathcal{O}(n-1)^{\oplus n}$.
This is wrong. See David Speyer's response for the correct expression.
Now for the questions.
**Q1**. I would like to look at the morphism $\mathbf{P}^r\longrightarrow \mathbf{P}^r$ given by $[x\_0:\ldots:x\_r]\mapsto [x\_0^n:\ldots:x\_0^r]$. This is a finite morphism and therefore it should be possible to apply GRR in finding an expression for $\pi\_\ast \mathcal{O}(m)$, right? Now, this is probably still very easy so I was wondering if there were any other results in this direction.
**Q2**. How would one do this without the Grothendieck-Riemann-Roch theorem? By "this", I also mean the above example for $\mathbf{P}^1$.
**Q3**. What is the geometric interpretation of this?
**Final note**. Since every vector bundle on $\mathbf{P}^1$ has a unique decomposition into twisted sheaves, we get a nice expression for $\pi\_\ast \mathcal{E}$ where $\mathcal{E}$ is a vector bundle on $\mathbf{P}^1$.
| https://mathoverflow.net/users/4333 | On finite endomorphisms of $\mathbf{P}^r$ | The claimed computation is still wrong. Let $m \equiv r \mod n$, with $0 \leq r < n$. Then the right answer is that
$$\pi\_\* \mathcal{O}(m) = \mathcal{O}( \lfloor (m+1)/n \rfloor-1)^{\oplus(n-r-1)} \oplus \mathcal{O}( \lceil (m+1)/n \rceil-1)^{\oplus(r+1)}.$$
Let $S$ be the source $\mathbb{P}^1$ and $T$ the target. As a general piece of advice, you should never identify two spaces when you don't have to. Here are three ways you could get this answer:
By Grothendieck-Riemmann-Roch
------------------------------
Let $L\_S$ be the line bundle $\mathcal{O}(1)$ on $S$ and $L\_T$ the line bundle $\mathcal{O}(1)$ on $T$. Let $H\_S$ and $H\_T$ be the hyperplane classes in $H^\*(S)$ and $H^\*(T)$. The chern character of $L\_S^{\otimes m}$ is $(1+H\_S)^m = 1+m H\_S$. The Todd classes of $S$ and $T$ are $1+H\_S$ and $1+H\_T$. So $\pi\_\* L\_S^{\otimes m}$ is something with chern character
$$(1+H\_T)^{-1} \pi\_\*\left( (1+m H\_S) (1+H\_S) \right) = (1+H\_T)^{-1} \pi\_\*\left( 1+(m+1) H\_S \right).$$
Note that $\pi\_\* 1 = n$ and $\pi\_\* H\_S = H\_T$. So we get
$$ch(\pi\_\* \mathcal{O}(m)) = (1+H\_T)^{-1} (n+(m+1) H\_T)=n + (m-n+1) H\_T.$$
(We know that $R^1 \pi\_\*$ vanishes because the map is finite.)
From the leading term, we see that $\pi\_\* \mathcal{O}(m)$ has rank $n$. It is not completely obvious that is torsion free. If we assume it is, then it must be of the form $\bigoplus \mathcal{O}(a\_i)$ for some $a\_1 + \cdots + a\_n$. We see from the above computation that $\sum a\_i = m-n+1$.
That's as far as we can get from GRR. Grothendieck-Riemann-Roch can only do the computation in $K$-theory, so we can't distinguish $\mathcal{O}(-1) \oplus \mathcal{O}(1)$ from $\mathcal{O}(0) \oplus \mathcal{O}(0)$.
Directly in K-theory
--------------------
The point of Grothendieck-Riemann-Roch is that it gives a commuting diagram
$$\begin{matrix} K^0(S) & \longrightarrow & H^\*(S) \\
\downarrow & & \downarrow \\
K^0(T) & \longrightarrow & H^\*(T). \end{matrix}$$
I usually find that it is just as easy to do my computations directly on the left hand side of the diagram. Let $p\_S$ and $p\_T$ be the class of the structure sheaf of a point on $S$ and $T$. We have the short exact sequence $0 \to \mathcal{O}(-1) \to \mathcal{O}(0) \to \mathcal{O}\_{\mathrm{pt}} \to 0$, so $p\_S = 1-L\_S^{-1}$ and $L\_S = 1+p\_S$. (Since $p\_S^2=0$.)
Clearly, $\pi\_\* p\_S = p\_T$. So
$$\pi\_\* \mathcal{O}(m) = \pi\_\* (1+p\_S)^m = \pi\_\* (1+m p\_S) = \pi\_\* 1 + m p\_T.$$
We can see that $\pi\_\* 1$ has rank $n$; say $\pi\_\* 1 = n+a p\_T$.
Let $\chi$ be pushforward to the $K$-theory of a point, better known as holomorphic Euler characteristic.
Since pushforward is functorial, the sequence of maps $S \to T \to \mathrm{pt}$ shows that $$\chi(\pi\_\* 1) = \chi(1) = 1$$
so $n+a=1$ and $a = -(n-1)$. We see that
$$\pi\_\* \mathcal{O}(m) = n + (m-n+1) p\_T.$$
By direct computation
---------------------
It is easy enough to do this example, and any toric example like it, directly from the definition of pushforward. As a bonus, this will tell us exactly which vector bundle we get, not just the $K$-class.
Let $S\_1 \cup S\_2$ be the open cover of $S$ where $S\_1 = \mathrm{Spec} \ k[s]$ and $S\_2 = \mathrm{Spec} \ k[s^{-1}]$. Similarly, define $T\_1$, $T\_2$, $k[t]$ and $k[t^{-1}]$.
Let $e$ be a generator for the free $k[s]$ module $\mathcal{O}(m)(S\_1)$. Then $s^m e$ is a generator of $\mathcal{O}(m)(S\_2)$. By definition, $\left( \pi\_\* \mathcal{O}(m) \right) (T\_1)$ is $\mathcal{O}(m)(S\_1)$ considered as a $k[t]$-module. As such, it has basis
$$ e,\ s e,\ s^2 e,\ \ldots s^{n-1} e. \quad (\*)$$
Similarly, $\left( \pi\_\* \mathcal{O}(m) \right) (T\_2)$ has basis
$$ s^m e,\ s^{m-1} e,\ \ldots, s^{m-n+1} e. \quad (\*\*)$$
Reorder the lists $(\*)$ and $(\*\*)$ so that corresponding elements have exponents of $s$ which are congruent modulo $n$. To keep notation simple, I'll do the case of $m=0$. So we pair off:
* $e$ and $e$
* $s e$ and $s^{-n+1} e = t^{-1} (s e)$
* $s^2 e$ and $s^{-n+2} e = t^{-2} (s e)$
and so forth.
There is one time that we pair $v$ with itself and $(n-1)$ times that we pair $v$ with $t^{-1} v$. So the transition matrix between our bases is diagonal with entries $(1,t^{-1}, t^{-1}, \ldots, t^{-1})$ and the vector bundle is $\mathcal{O}(0) \oplus \mathcal{O}(-1)^{n-1}$.
For general $m$, if I didn't make any errors, we get the formula at the beginning of the post.
| 5 | https://mathoverflow.net/users/297 | 19108 | 12,721 |
https://mathoverflow.net/questions/19105 | 1 | This is a slight reformulation of exercise II.5.9.(c) in Hartshorne's "Algebraic Geometry" which I don't understand.
>
> Let $K$ be a field and $S=K[X\_0,\ldots,X\_n]$ a graded ring. Set $X=Proj(S)$ and let $M$ be a graded $S$-module. The functors $\Gamma\_\*$ definied by
> $$
> \Gamma\_\*(\mathcal{F})=\bigoplus\_{n\in\mathbb{Z}} (\mathcal{F}(n))(X)
> $$
> and $~\widetilde{\phantom{\cdot}}~$ (the "graded associated sheaf functor", see Hartshorne II.5. page 116 for a definition)
> induce an **equivalence of categories** between the category $\mathcal{A}$ of quasi-finitely generated (i.e. in relation to a finitely generated module) graded $S$-modules modulo a certain equivalence relation $\approx$ and the category $\mathcal{B}$ of coherent $\mathcal{O}\_X$-modules. The equivalence relation is: $M\approx N$ if there is an integer $d$ such that $\oplus\_{k\geq d}M\_k\cong\oplus\_{k\geq d}N\_k$.
>
>
>
I don't know what an "equivalence of categories" is in this context. Formally an "equivalence of categories" means in particular that there are isomorphisms $$\hom\_\mathcal{A}(M,N)\cong \hom\_\mathcal{B}(\widetilde{M},\widetilde{N})$$ and $$\hom\_\mathcal{B}(Y,Z)\cong \hom\_\mathcal{A}(\Gamma\_\*(Y),\Gamma\_\*(Z))$$
of **sets**. This is my problem: How is the sheaf $\mathcal{H}om\_\mathcal{B}(Y,Z)$ considered as a **set**? Perhaps it should be $\Gamma\_\*(\mathcal{H}om\_\mathcal{B}(Y,Z))\cong \hom\_\mathcal{A}(\Gamma\_\*(Y),\Gamma\_\*(Z))$?
| https://mathoverflow.net/users/2625 | Question on an exercise in Hartshorne: Equivalence of categories | The homomorphisms in the category of sheaves are not sheaves themselves. The hom sheaves have the data of things that are only homomorphisms over open subsets. So if $Y,Z$ are coherent $\mathcal{O}\_X$-modules and you are looking for $\mathcal{O}\_X$-module homomorphisms, you don't actually get $\mathcal{H}om(X,Y)$, what you actually get are the global sections only, because these are the only homomorphisms that are defined on the whole space, and so the only actual homomorphisms in the category of sheaves.
| 4 | https://mathoverflow.net/users/622 | 19110 | 12,723 |
https://mathoverflow.net/questions/19104 | 5 | Is there a space such that it doesn't have any deformation but its space of first order infinitesimal deformations is non trivial?
| https://mathoverflow.net/users/4821 | Deformations for complex space germs | Such germs of spaces exist. See Section 7.6 of Ravi Vakil's paper *[Murphy's law in Algebraic Geometry](http://arxiv.org/abs/math/0411469)*.
| 5 | https://mathoverflow.net/users/297 | 19113 | 12,726 |
https://mathoverflow.net/questions/19119 | 8 | Let $\mathcal{F}$ and $\mathcal{G}$ be two sheaves (of abelian groups) on a topological space $X$ such that $\mathcal{G}(U)$ is a subgroup of $\mathcal{F}(U)$ for every open set $U$ in $X$. The sheaf associated to the presheaf $P(\mathcal{F}/\mathcal{G})$ defined by
$$
U\mapsto \mathcal{F}(U)/\mathcal{G}(U)
$$
is called the quotient sheaf $\mathcal{F}/\mathcal{G}$. The associated sheaf functor is left adjoint to the inclusion functor, so it commutes with colimits and in particular with quotients. My question is: Why must one sheafify the presheaf $P(\mathcal{F}/\mathcal{G})$ then?
| https://mathoverflow.net/users/2625 | Why must one sheafify quotients of sheaves? | For presheaves (of sets or groups) we know what this particular (or any) colimit operation is: apply the operation objectiwise (for each $U$). Now the sheafification preserves colimits, hence we apply sheafification to a colimit cocone on presheaves to obtain a colimit cocone in sheaves. Doing sheafification to the presheaves which are alerady sheaves does nothing to them, but it, by the right exactness, does the correct thing to the colimit. This proves that the sheafification following the colimit in presheaves is the correct way to compute the colimit, and in that we did use the right exactness of the sheafification essentially. The fact that it is necessary does not follow from the general nonsense as there are both examples where we accidentally do not need a sheafification step and those where we do need. For the limit constructions on sheaves we never need the sheafification
because the embedding of the sheaves into presheaves is left exact hence we can simply compute the limits in presheaves. It seems you had somehow an opposite impression.
| 16 | https://mathoverflow.net/users/35833 | 19122 | 12,731 |
https://mathoverflow.net/questions/18766 | 26 | Let $\approx$ be the binary relation on the class of finitely generated groups
such that $G \approx H$ iff $G$ and $H$ have isomorphic (unlabeled nondirected)
Cayley graphs with respect to suitably chosen finite generating sets. Is $\approx$ an equivalence relation?
| https://mathoverflow.net/users/4706 | Cayley graphs of finitely generated groups | The answer is no, as expected. The following proof is "joint work" with L. Scheele.
Consider $G=\mathbb{Z}$, $K=D\_\infty$ and $H:=\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$. Then $G \approx K$ and $K\approx H$, but $G \not\approx H.$
Indeed, the Cayley graph associated to $\{-1,1\}$ for G and the Cayley graph associated to $\{s,t\}$ where $D\_\infty=\langle s,t: s^2=t^2=1 \rangle$ are clearly isometric.
Similarly, the Cayley graph associated to $\{s,st,ts\}$ for $K$ and the graph associated to $\{(0,1),(-1,0),(1,0)\}$ for $H$ are isometric.
However, let $S$ be some symmetric generating set for $G$. Then $S\_k$, the set of vertices which have distance precisely $k\geq 1$ from the identity, has even cardinality because $S\_k$ is invariant under the mapping $x \mapsto -x$ and doesn't contain 0.
Now let $T$ be some symmetric generating set for $H$. Let $k\_0$ be the distance of $(0,1)$ from the identity in the associated graph. Then $T\_{k\_0}$, the set of vertices which have distance precisely $k\_0$ from the identity, has odd cardinality. Indeed, it is invariant under the mapping $(x,y) \mapsto (-x,-y)$ since $T$ is assumed to be symmetric. But $(0,1)$ is a fixed point.
| 18 | https://mathoverflow.net/users/3380 | 19124 | 12,733 |
https://mathoverflow.net/questions/19132 | 5 | Maybe this is an elementary question.
Suppose that $U$ is a non-principal $\kappa$-complete ultrafilter on $\kappa$ and consider the standard ultrapower $M\cong \textrm{Ult}\_U(V)$ along with the corresponding elementary embedding $j\_U : V \rightarrow M$\,.
We know that if $U$ is in addition normal, then every element $y\in M$ can be written in the form $j(f)(\kappa)$ (more precisely, if $y=[f]\_U \in M$ then $y=j(f)(\kappa)$, which follows from the fact that $[id]\_U=\kappa$).
Of course, we can always pick $U$ to be normal, but the question is what happens in the case that we don't. More precisely:
Assume that in $M$, $[g]=\kappa<[id]\_U$, for some $g:\kappa\rightarrow \kappa$. Does it follow that for every $y\in M$, $y=j(f)(\kappa)$ for some function $f$ on $\kappa$?
My guess is that the desired would follow if can pick the function $g$ that represents $\kappa$ to be 1-1 and such that $\textrm{range}(g)\in U$. Is this true? Can we do this?
| https://mathoverflow.net/users/4826 | Representation of elements in ultrapowers | What you get is that if j:V to M is the ultrapower by any ultrafilter U on any set X, then every element of M has the form j(f)([id]). You can prove this by building an isomorphism from the ultrapower to the sets of this form. This way of thinking is also known as "seed theory".
**Theorem.** Suppose that j:V to M is an elementary embedding of the universe V into M. Then j is the ultrapower map by a measure on a set if and only if there is some s in M such that every element of M has the form j(f)(s).
That is, the ultrapower embeddings are precisely the embeddings whose target is generated by a single element.
Proof. If j is the ultrapower by U on X, then let s=[id], and argue that [f]\_U is j(f)(s). Conversely, if M = { j(f)(s) | f in V }, where we assume that f is a function on some set X such that a ∈ j(X), then define the measure U by A ∈ U iff s ∈ j(A). This is a κ-complete ultrafilter on P(X). One can show that Ult(V,U) is isomorphic to M, by mapping [f]\_U to j(f)(s). QED
**Theorem.** If U is a κ complete ultrafilter on κ with ultrapower embedding j:V to M, then every element of M has the form j(f)(κ) if and only if U is isomorphic to a normal measure.
Proof. You know the backwards implication. For the forward implication, suppose that every element of M has form j(f)(κ). In particular, β = j(f)(κ), where β = [id]\_U is the seed for U. But also, κ = j(g)(β), where g(α) is the smallest ξ for which f(ξ)=α. Let μ = { X subset κ | κ ∈ j(X) } be the induced normal measure. Note that X in μ iff j(g)(β) in j(X) iff β in j(g-1X) iff g-1X in U. So μ is Rudin-Kiesler below U. Also, U is Rudin-Kiesler below μ since X in U iff f-1X in μ. So μ and U are isomorphic.QED
One may illustrate the situation with product measures. Suppose that U is normal. The product measure UxU is isomorphic to the two-step iteration, where j\_0:V to M is the ultrapower by U, and h:M to N is the ultrapower in M by j\_0(U). Every element of M has the form j\_0(f)(κ), and every element of N has the form h(g)(κ1), where κ1 = j\_0(κ). If j is the composition of j\_0 and h, then j:V to N and every element of N has the form j(f)(κ, κ1). If one only looks at j(f)(κ) inside N, then you will only get ran(h), which is isomorphic to M, but not all of N. So this would be a counterexample to what you asked about.
| 7 | https://mathoverflow.net/users/1946 | 19135 | 12,740 |
https://mathoverflow.net/questions/19142 | 1 | Let R be a commutative ring with 1.
An R-module K has the 'S' property if K/T = K implies that the submodule T is trivial.
By Fitting's Lemma any Noetherian module has the 'S' property. There exist non-Noetherian modules with this property. For example the infinite product of Z\_{2}xZ\_{3}xZ\_{5}x... running over all of the primes has the 'S' property, but is not Noetherian.
I am curious if there is a characterization of these kinds of modules out there.
| https://mathoverflow.net/users/4828 | Characterization of a certain class of modules-broader than Noetherian | These are called **Hopfian modules**. I didn't see any particularly exciting general characterization, but there are several special case characterizations (that show up easily in a google or mathscinet search). There are also several papers devoted to giving "interesting" examples.
An exercise in Lam's Lectures on Modules and Rings asks one to prove that every finitely generated module over a commutative ring is Hopfian (so if the ring is not-noetherian, this is a generalization).
| 5 | https://mathoverflow.net/users/3710 | 19144 | 12,746 |
https://mathoverflow.net/questions/19063 | 7 | Shapovalov and Jantzen showed us how to construct a nice inner product on finite dimensional representations of a semi-simple Lie algebra, by simply giving the highest weight vector inner product 1 with itself and making the upper and lower halves adjoint.
The result I need is an extension of this to tensor products. Roughly, I would like a statement like:
>
> There is a unique system of $U\_q(\mathfrak{g})$-invariant Hermitian inner products on all tensor products $V\_{\lambda\_1}\otimes \cdots \otimes V\_{\lambda\_\ell}$ such that
>
>
> 1. On $V\_\lambda$, it is the Shapovalov form.
> 2. The action of $E\_i$ and $F\_i$ are biadjoint (up to some powers of $q$).
> 3. For any $j<\ell$, the natural map $V\_{\lambda\_1}\otimes\cdots\otimes V\_{\lambda\_j}\hookrightarrow V\_{\lambda\_1}\otimes \cdots \otimes V\_{\lambda\_\ell}$ is an isometric embedding.
>
>
>
I said to myself, "Self, it would be silly to post this question on MathOverflow. You are in a math library, just feet away from Lusztig's book. Surely it is in there." However, I've had no luck finding it in Lusztig's book, which is sadly lacking in index. Is this actually written down anywhere?
**EDIT**: Jim asks for more motivation. I feel like this is the sort of question where motivation will not be very helpful in actually finding an answer, but there's no harm in saying a little (and it will allow me to put off real work).
One of the foundational principles of categorification is that things with nice categorifications have nice inner products (since Grothendieck groups have a nice inner product given by Euler characteristic of the Ext's between objects). I'm working right now on categorifying tensor products of representations, so it would be rather convenient for me to find some earlier references that used this form.
| https://mathoverflow.net/users/66 | Reference for the existence of a Shapovalov-type form on the tensor product of integrable modules | I know a couple of ways to get a Shapovalov type form on a tensor product. The details of what I say depends on the exact conventions you use for quantum groups. I will follow Chari and Pressley's book.
The first method is to alter the adjoint slightly. If you choose a \* involution that is also a coalgebra automorphism, you can just take the form on a tensor product to be the product of the form on each factor, and the result is contravariant with respect to \*. There is a unique such involution up to some fairly trivial modifications (like multiplying $E\_i$ by $z$ and $F\_i$ by $z^{-1}$). It is given by:
$$
\*E\_i = F\_i K\_i, \quad \*F\_i=K\_i^{-1}E\_i, \quad \*K\_i=K\_i,
$$
The resulting forms are Hermitian if $q$ is taken to be real, and will certainly satisfy your conditions 1) ad 3). Since the $K\_i$s only act on weight vectors as powers of $q$, it almost satisfies 2).
The second method is in case you really want \* to interchange $E\_i$ with exactly $F\_i$. This is roughly contained in this <http://www.ams.org/mathscinet-getitem?mr=1470857> paper by Wenzl, which I actually originally looked at when it was suggested in an answer to one of your previous questions.
It is absolutely essential that a \* involution be an algebra-antiautomorphism. However, if it is a coalgebra anti-automorphism instead of a coalgebra automorphism there is a work around to get a form on a tensor product. There is again an essentially unique such involution, given by
$$
\*E\_i=F\_i, \quad \*F\_i=E\_i, \quad \*K\_i=K\_i^{-1}, \quad \*q=q^{-1}.
$$
Note that $q$ is inverted, so for this form one should think of $q$ as being a complex number of the unit circle. By the same argument as you use to get the Shapovalov form, then is a unique sesquilinear \*-contravariant form on each irreducible representation $V\_\lambda$, up to overall rescaling.
To get a form on $V\_\lambda \otimes V\_\mu$, one should define
$$(v\_1 \otimes w\_1, v\_2 \otimes w\_2)$$
to be the product of the form on each factor applied to $v\_1 \otimes w\_1$ and $R( v\_2 \otimes w\_2)$, where $R$ is the universal $R$ matrix. It is then straightforward to see that the result is \*-contravariant, using the fact that $R \Delta(a) R^{-1} =\Delta^{op}(a).$
If you want to work with a larger tensor product, I believe you replace $R$ by the unique endomorphism $E$ on $\otimes\_k V\_{\lambda\_k}$ such that $w\_0 \circ E$ is the braid group element $T\_{w\_0}$ which reverses the order of the tensor factors, using the minimal possible number of positive crossings. Here $w\_0$ is the symmetric group element that reverses the order of the the tensor factors.
The resulting form is \*-contravariant, but is not Hermitian. In Wenzl's paper he discusses how to fix this.
Now 1) and 2) on your wish list hold. As for 3): It is clear from standard formulas for the $R$-matrix (e.g. Chari-Pressley Theorem 8.3.9) that $R$ acts on a vector of the form $b\_\lambda \otimes c \in V\_\lambda \otimes V\_\mu$ as multiplication by $q^{(\lambda, wt(c))}$. Thus if you embed $V\_\mu$ into $V\_\lambda \otimes V\_\mu$ as $w \rightarrow b\_\lambda \otimes w$, the result is isometric up to an overall scaling by a power of $q$. This extends to the type of embedding you want (up to scaling by powers of $q$), only with the order reversed. I don't seem to understand what happen when you embed $V\_\lambda$ is $V\_\lambda \otimes V\_\mu$, which confuses me, and I don't see your exact embeddings.
| 6 | https://mathoverflow.net/users/1799 | 19145 | 12,747 |
https://mathoverflow.net/questions/19127 | 93 | I heard the following two questions recently from [Carl Mummert](http://users.marshall.edu/~mummertc/), who encouraged me to spread them around. Part of his motivation for the questions was to give the subject of computable model theory some traction on complete metric spaces, by considering the countable objects as stand-ins for the full spaces, to the extent that they are able to do so.
**Question 1.** Is there a countable subset $D$ of the real plane $\mathbb{R}^2$ that is dense and has the property that the Euclidean distance $d(x,y)$ is a rational number for all $x,y\in D$?
The one dimensional analogue of this question has an easy affirmative answer, since the rationals $\mathbb{Q}$ sits densely in $\mathbb{R}$ and the distance between any two rationals is rational.
**Question 2.** More generally, does every separable complete metric space have a countable dense set $D$ with all distances between elements of $D$ rational? [Edit: Tom Leinster has pointed out that if the space has only two points, at irrational distance, this fails. So let us consider the case of connected spaces, generalizing the situation of Question 1.]
If one is willing to change to an equivalent metric (giving rise to the same topology), then the answer to Question 1 is Yes, since the rational plane $\mathbb{Q}\times\mathbb{Q}$ is dense in the real plane $\mathbb{R}\times\mathbb{R}$, and has all rational distances under the Manhattan metric, which gives rise to the same topology. Is the answer to the correspondingly weakened version of Question 2 also affirmative, if one is willing to change the metric?
Note that one cannot find an equivalent metric such that *all* distances in $\mathbb{R}^2$ become rational, since omitted values in the distance function lead to disconnectivity in the space. This is why the questions only seek to find a dense subset with the rational condition.
The question seems related to the question of whether it is possible to find large non-linear arrangements of points in the plane with all pair-wise distances being integers. For example, this is true of the integers $\mathbb{Z}$ sitting inside $\mathbb{R}$, but can one find a 2 dimensional analogue of this? Clearly, some small arrangements (triangles, etc.) are possible, but I am given to understand that there is a finite upper bound on the size of such arrangements. What is the precise statement about this that is known?
| https://mathoverflow.net/users/1946 | Is there a dense subset of the real plane with all pairwise distances rational? | Let me answer Question 2.
Strong version: no. Consider $[0,1]$ with distance $d(x,y)=|x-y|^{1/3}$. There is no even a triple of points with rational distances - otherwise there would be a nonzero rational solution of $x^3+y^3=z^3$.
Weak version: yes. Let $(X,d)$ be the space in question. Construct sets $S\_1\subset S\_2\subset\dots$ such that each $S\_k$ is a maximal $(2^{-k})$-separated net in $X$. Let $S$ be the union of these nets; then $S$ is countable and dense in $X$.
Now construct the following metric graph on $S$. For every $k$, connect every pair of points $x,y\in S\_k$ by an edge whose length is $(1-10^{-k})d(x,y)$ rounded down to a multiple of $10^{-2k}$. The new distance $d'$ on $S$ is the induced length distance in this graph. It is easy to see that the edges outside $S\_k$ do not affect the distances in $S\_k$, hence all these distances are rational (multiples of $10^{-2k}$). The new metric $d'$ on $S$ satisfies $\frac12d\le d'\le d$, hence the completion of $(S,d')$ is the same set $X$ with an equivalent metric.
**UPDATE**.
Here is a more detailed description without the term "metric graph".
For each $k$, define a function $f\_k:\mathbb R\_+\to\mathbb R\_+$ by
$$
f\_k(t) = 10^{-2k}\left\lfloor 10^{2k}(1-10^{-k})t \right\rfloor .
$$
The actual form of $f\_k$ does not matter, we only need the following properties:
* $f\_k$ takes only rational values with bounded denominators (by $10^{-k}$).
* Let $a\_k$ and $b\_k$ denote the infimum and the supremum of $f\_k(t)/t$ over the set $\{t\ge 2^{-k}\}$. Then $\frac12\le a\_k\le b\_k\le a\_{k+1}\le 1$ for all $k$. (Indeed, we have $1-2\cdot10^k\le a\_k\le b\_k\le 1-10^k$.)
For every $x,y\in S\_k$, define $\ell(x,y)=f\_k(d(x,y))$ where $k=k(x,y)$ is the minimum number such that $x,y\in S\_k$. Note that
$$
a\_k d(x,y) \le \ell(x,y) \le b\_k d(x,y)
$$
for all such pairs $x,y$, since $S\_k$ is a $(2^{-k})$-separated set. For a finite sequence $x\_0,x\_1,\dots,x\_n\in S$ define
$$
\ell(x\_0,x\_1,\dots,x\_n) = \sum\_{i=1}^n \ell(x\_{i-1},x\_i) .
$$
I will refer to this expression as the $\ell$-*length* of the sequence $x\_0,\dots,x\_n$. Define
$$
d'(x,y) = \inf\{ \ell(x\_0,x\_1,\dots,x\_n) \}
$$
where the infimum is taken over all finite sequences $x\_0,x\_1,\dots,x\_n$ in $S$ such that $x\_0=x$ and $x\_n=y$. Clearly $d'$ is a metric and $\frac12d\le d'\le d$. It remains to show that $d'$ takes only rational values.
**Lemma**: If $x,y\in S\_k$, then $d'(x,y)$ equals the infimum of $\ell$-lengths of sequences contained in $S\_k$.
*Proof*: Consider any sequence $x\_0,\dots,x\_n$ in $S$ such that $x\_0=x$ and $y\_0=y$. Remove all points that do not belong to $S\_k$ from this sequence. I claim that the $\ell$-length became shorter. Indeed, it suffices to prove that
$$
\ell(x\_r,x\_s) \le \ell(x\_r,x\_{r+1},\dots,x\_{s-1},x\_s)
$$
if $x\_r$ and $x\_s$ are in $S\_k$ and the intermediate points are not. By the second property of the functions $f\_k$, the left-hand side is bounded above by $b\_k d(x\_r,x\_s)$ and every term $\ell(x\_i,x\_{i+1})$ in the right-hand side is bounded below by $b\_k d(x\_i,x\_{i+1})$. So it suffices to prove that
$$
b\_k d(x\_r,x\_s) \le b\_k\sum\_{i=r}^{s-1} d(x\_i,x\_{i+1}),
$$
and this is a triangle inequality multiplied by $b\_k$. Q.E.D.
All $\ell$-lengths of sequences in $S\_k$ are multiples of some fixed rational number (namely $10^{-2k}$). Hence $d'(x,y)$ is a multiple of the same number if $x,y\in S\_k$. Thus all values of $d'$ are rational.
| 62 | https://mathoverflow.net/users/4354 | 19153 | 12,751 |
https://mathoverflow.net/questions/19118 | 9 | I guess this question only requires standard knowledge, but I'm a bit rusty with highest weight theory. I'm trying to catch up, but maybe I don't need the theory in full generality.
**Background**
Let $V$ be a Euclidean space of dimension $n$, and consider the representations of the group $G = O(V) \cong O(n)$. If I'm not wrong we can decompose
$$
\operatorname{Sym}^2 V = W \oplus Z,
$$
where $Z$ is the trivial one-dimensional representation and $W$ is an irreducible representation of $G$.
**More background**
This is how I can see the above decomposition. I don't know if it is of any use for the question itself, so feel free to skip it.
It is enough to decompose $\operatorname{Sym}^2 V^{\*}$, that is, degree $2$ homogeneous polynomials on $V$. Degree $n$ homogeneous polynomials are a representation of $G$ via
$$
A.f(x) = f(Ax),
$$
where $A$ is an orthogonal matrix and $f \in \operatorname{Sym}^n V^{\*}$.
Now for any $n$ we have a $G$-morphism $f : \operatorname{Sym}^n V^{\*} \to \operatorname{Sym}^{n + 2} V^{\*}$ which is given by multiplication by $x^2 = x\_1^2 + \cdots + x\_n^2$. This exhibits $\operatorname{Sym}^n V^{\*}$ as a subrepresentation of $\operatorname{Sym}^{n + 2} V^{\*}$.
The complement is easily found. The invariant scalar product on $\operatorname{Sym}^n V^{\*}$ is given by $(f, g) = f(D)g(x)$, where $D$ is the derivation operator. From this one finds that the adjoint of $f$ is the laplacian $\Delta$.
So one can decompose
$$
\operatorname{Sym}^{n + 2} V^{\*} = \operatorname{Sym}^{n} V^{\*} \oplus \mathcal{H}\_n,
$$
where $\mathcal{H}\_n$ is the space of harmonic homogeneous polynomials of degree $n$. If I recall well, $\mathcal{H}\_n$ is irreducible.
The decomposition which interests me is then
$$
\operatorname{Sym}^{2} V^{\*} = \mathcal{H}\_2 \oplus \mathcal{H}\_0.
$$
**Problem**
I'd like to understand how to decompose the tensor products of $W$; in particular
>
> What is the decomposition of $W \otimes W$ and $W \otimes W \otimes W$ into irreducible representations of $G$?
>
>
>
| https://mathoverflow.net/users/828 | Decomposing a tensor product | For $SO(n)$ a calculation using LiE gives:
(using partition notation so $W$ is [2])
and assuming $n$ is not small
For $W\otimes W$, [4],[3,1],[2,2],[2],[1,1],[]
(all with multiplicity one)
and for $W\otimes W\otimes W$,
1.[6] 2.[5,1] 3.[4,2] 1.[3,3] 1.[4,1,1] 2.[3,2,1] 1.[2,2,2] 3.[4] 6.[3,1] 2.[2,2] 3.[2,1,1] 6.[2] 3.[1,1] 1.[]
The same works for $Sp(n)$ by taking conjugate partitions.
There is also a relationship with $SL(n)$.
This is taking your question at face value. If it is understanding you're after instead then the best approach is to use crystal graphs.
The notation I have used denotes a representation by a partition. I have put $m.$ in front to denote multiplicity is $m$. A partition is $[a\_1,a\_2,a\_3,...]$ where $a\_i\ge a\_j$ if $i$ less than $j$. To convert to a highest weight vector add the appropriate number of $0$s to the end.
Then take $[a\_1-a\_2,a\_2-a\_3,a\_3-a\_4,...]$. This gives a dominant integral weight. The fundamental weights are the partitions $[1,,,,1]$. If this has length $k$ this corresponds to the $k$-th exterior power of the vector representation (provided $2k-1$ less than $n$).
In particular the trivial representation is $[]$, the vector representation $V$ is $[1]$, the exterior square of $V$ is $[1,1]$, the symmetric square is $[2]+[]$.
For the $k$-th tensor power of $W$ you will see partitions of $2k-2p$ for $0\le p\le k$ only and it remains to determine the multiplicities (possibly $0$). For $SL(n)$ just take the partitions of $2k$ (with their multiplicities) and ignore the rest.
| 4 | https://mathoverflow.net/users/3992 | 19154 | 12,752 |
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